[DISCRETE-TIME](#page-11-0) SYSTEM ANALYSIS USING THE z-TRANSFORM

← Back to LINEAR SYSTEMS AND SIGNALS Overview \frac{2s+5}{s^2+5s+6}

$$\n(b)$$

\frac{3s+5}{s^2+4s+13}

$$\n(c)$$

\frac{(s+1)^2}{s^2-s-6}

$$\n(d)$$

\frac{5}{s^2(s+2)}

$$**Figure P4.1-3** (e)$$

\frac{2s+1}{(s+1)(s^2+2s+2)}

$$\n(f)$$

\frac{s+2}{s(s+1)^2}

$$\n(g)$$

\frac{1}{(s+1)(s+2)^4}

$$\n(h)$$

\frac{s+1}{s(s+2)^2(s^2+4s+5)}

$$(i)$$

\frac{s(s+2)^2(s^2+4s+5)}{(s+1)^2(s^2+2s+5)}

$$- **4.2-1** Suppose a CT signal *x*(*t*)=2[*u*(*t* −2)−*u*(*t* +1)] has a transform *X*(*s*). - (a) If *^Y*a(*s*) ⁼ *^e*−5*^s sX ^s* ⁺ ¹ 2 , determine and sketch the corresponding signal *y*a(*t*). - (b) If *^Y*b(*s*) ⁼ ²−*^s sX*(*s* −2), determine and sketch the corresponding signal *y*b(*t*). - **4.2-2** Find the Laplace transforms of the following functions using only Table 4.1 and the time-shifting property (if needed) of the unilateral Laplace transform: - (a) *u*(*t*) −*u*(*t* −1) - (b) *^e*−(*t*−τ )*u*(*^t* ⁻τ ) - (c) *e*−(*t*−τ )*u*(*t*) - (d) *e*−*^t u*(*t* −τ ) - (e) *te*−*^t u*(*t* −τ ) - (f) sin[ω0(*t* −τ )]*u*(*t* −τ ) - (g) sin[ω0(*t* −τ )]*u*(*t*) - (h) sin ω0*t u*(*t* −τ ) - (i) *t*sin(*t*)*u*(*t*) - (j) (1−*t*) cos(*t* −1)*u*(*t* −1) - **4.2-3** Using only Table 4.1 and the time-shifting property, determine the Laplace transform of the signals in Fig. P4.1-3. [*Hint:* See Sec. 1.4 for discussion of expressing such signals analytically.] - **4.2-4** Prove the frequency-differentiation property, ⁻*tx*(*t*) ⇐⇒ *^d dsX*(*s*). This property holds for both the unilateral and bilateral Laplace transforms. - **4.2-5** Consider the signal *^x*(*t*) ⁼ *te*−2(*t*−3) *u*(*t*−2). - (a) Determine the *unilateral* Laplace transform *X*u(*s*) = *L*^u {*x*(*t*)}. - (b) Determine the *bilateral* Laplace transform *X*(*s*) = *L*{*x*(*t*)}. - **4.2-6** Consider the signals *x*(*t*) and *y*(*t*), as shown in Fig. P4.2-6. - (a) Using the definition, compute *X*(*s*), the bilateral Laplace transform of *x*(*t*). - (b) Using Laplace transform properties, express *Y*(*s*), the bilateral Laplace transform of *y*(*t*), as a function of *X*(*s*), the bilateral Laplace transform of *x*(*t*). Simplify as much as possible without substituting your answer from part (a). - **4.2-7** Find the inverse Laplace transforms of the following functions: (a)$$

\frac{(2s+5)e^{-2s}}{s^2+5s+6}

$$(b)$$

\frac{se^{-3s}+2}{s^2+2s+2}

$$(c)$$

\frac{e^{-(s-1)}+3}{2(2s+5)}

$$$$

\begin{array}{c}\n\text{(c)} \quad s^2 - 2s + 5 \ \text{(d)} \quad \frac{e^{-s} + e^{-2s} + 1}{s^2 + 3s + 2}\n\end{array}

$$- **4.2-8** Using ROC σ > 0, determine the inverse Laplace transform of *^X*(*s*) ⁼ *^s*−¹ *^d ds e*−2*^s s* . - **4.2-9** The Laplace transform of a causal periodic signal can be determined from the knowledge of the Laplace transform of its first cycle (period). - (a) If the Laplace transform of *x*(*t*) in Fig. P4.2-9a is *X*(*s*), then show that *G*(*s*), the Laplace transform of *g*(*t*) (Fig. P4.2-9b), is$$

G(s) = \frac{X(s)}{1 - e^{-sT_0}} \qquad \text{Re}, s > 0

- (b) Use this result to find the Laplace transform of the signal *p*(*t*) illustrated in Fig. P4.2-9c. - **4.2-10** Starting only with the fact that δ(*t*) ⇐⇒ 1, build pairs 2 through 10b in Table 4.1, using various properties of the Laplace transform. #### **Figure P4.2-9** - **4.2-11** (a) Find the Laplace transform of the pulses in Fig. 4.2 by using only the time-different iation property, the time-shifting property, and the fact that δ(*t*) ⇐⇒ 1. - (b) In Ex. 4.9, the Laplace transform of *x*(*t*) is found by finding the Laplace transform of *d*2*x*/*dt*2. Find the Laplace transform of *x*(*t*) in that example by finding the Laplace transform of *dx*/*dt* and using Table 4.1, if necessary. - **4.2-12** Determine the inverse unilateral Laplace transform of

X(s) = \frac{1}{e^{s+3}} \frac{s^2}{(s+1)(s+2)}

$$- **4.2-13** Since 13 is such a lucky number, determine the inverse Laplace transform of *X*(*s*) = 1/(*s* +1)¹³ given region of convergence σ > −1. [*Hint:* What is the *n*th derivative of 1/(*s* +*a*)?] - **4.2-14** It is difficult to compute the Laplace transform *X*(*s*) of signal$$

x(t) = \frac{1}{t}u(t)

$$by using direct integration. Instead, properties provide a simpler method. - (a) Use Laplace transform properties to express the Laplace transform of *tx*(*t*) in terms of the unknown quantity *X*(*s*). - (b) Use the definition to determine the Laplace transform of *y*(*t*) = *tx*(*t*). - (c) Solve for *X*(*s*) by using the two pieces from **()**(a) and **()**(b). Simplify your answer. - **4.3-1** Use the Laplace transform to solve the following differential equations: - (a) (*D*² + 3*D* + 2)*y*(*t*) = *Dx*(*t*) if *y*(0−) = *y*˙(0−) = 0 and *x*(*t*) = *u*(*t*) - (b) (*D*² + 4*D* + 4)*y*(*t*) = (*D* + 1)*x*(*t*) if *y*(0−) = 2, *^y*˙(0−) ⁼ 1 and *^x*(*t*) ⁼ *^e*−*^t u*(*t*) - (c) (*D*² +6*D*+25)*y*(*t*) = (*D*+2)*x*(*t*) if *y*(0−) = *y*˙(0−) = 1 and *x*(*t*) = 25*u*(*t*) - **4.3-2** Solve the differential equations in Prob. 4.3-1 using the Laplace transform. In each case determine the zero-input and zero-state components of the solution. - **4.3-3** Consider a causal LTIC system described by the differential equation$$

2\dot{y}(t) + 6y(t) = \dot{x}(t) - 4x(t)

$$- (a) Using transform-domain techniques, determine the ZIR *y*zir(*t*) if *y*(0−) = −3. - (b) Using transform-domain techniques, determine the ZSR *y*zsr(*t*) to the input *x*(*t*) = *e*δ(*t* −π ). - **4.3-4** Consider a causal LTIC system described by the differential equation$$

\ddot{y}(t) + 3\dot{y}(t) + 2y(t) = 2\dot{x}(t) - x(t)

$$- (a) Using transform-domain techniques, determine the ZIR *y*zir(*t*) if *y*˙(0−) = 2 and *y*(0−) = −3. - (b) Using transform-domain techniques, determine the ZSR *y*zsr(*t*) to the input *x*(*t*) = *u*(*t*). - **4.3-5** Solve the following simultaneous differential equations using the Laplace transform, assuming all initial conditions to be zero and the input *x*(*t*) = *u*(*t*): - (a) (*D*+3)*y*1(*t*)−2*y*2(*t*) = *x*(*t*) −2*y*1(*t*) +(2*D*+4)*y*2(*t*) = 0 - (b) (*D*+2)*y*1(*t*)−(*D*+1)*y*2(*t*) = 0 −(*D*+1)*y*1(*t*) +(2*D*+1)*y*2(*t*) = *x*(*t*) Determine the transfer functions relating outputs *y*1(*t*) and *y*2(*t*) to the input *x*(*t*). - **4.3-6** Consider a causal LTIC system described by *y*˙(*t*) +2*y*(*t*) = ˙*x*(*t*). - (a) Determine the transfer function *H*(*s*) for this system. - (b) Using your result from part (a), determine the impulse response *h*(*t*) for this system. - (c) Using Laplace transform techniques, determine the output *y*(*t*) if the input is *x*(*t*) = *e*−*^t ^u*(*t*) and *^y*(0−) ⁼ ^√ 2. - **4.3-7** Repeat Prob. 4.3-6 for a causal LTIC system described by 3*y*(*t*) + ˙*y*(*t*)+ ˙*x*(*t*) = 0. - **4.3-8** For the circuit in Fig. P4.3-8, the switch is in the open position for a long time before *t* = 0, when it is closed instantaneously. - (a) Write loop equations (in time domain) for *t* ≥ 0. - (b) Solve for *y*1(*t*) and *y*2(*t*) by taking the Laplace transform of loop equations found in part (a). **4.3-9** For each of the systems described by the following differential equations, find the system transfer function: (a)$$

\frac{d^2y(t)}{dt^2} + 11\frac{dy(t)}{dt} + 24y(t) = 5\frac{dx(t)}{dt} + 3x(t)

$$\n(b)$$

\frac{d^3y(t)}{dt^3} + 6\frac{d^2y(t)}{dt^2} - 11\frac{dy(t)}{dt} + 6y(t)

$$$$

= 3\frac{d^2x(t)}{dt^2} + 7\frac{dx(t)}{dt} + 5x(t)

$$\n(c)$$

\frac{d^4y(t)}{dt^4} + 4\frac{dy(t)}{dt} = 3\frac{dx(t)}{dt} + 2x(t)

$$\n(d)$$

\frac{d^2y(t)}{dt^2} - y(t) = \frac{dx(t)}{dt} - x(t)

$$**4.3-10** For each of the systems specified by the following transfer functions, find the differential equation relating the output *y*(*t*) to the input *x*(*t*), assuming that the systems are controllable and observable: (a)$$

H(s) = \frac{s+5}{s^2 + 3s + 8}

\n(b) $H(s) = \frac{s^2 + 3s + 5}{s^3 + 8s^2 + 5s + 7}$ \n(c) $H(s) = \frac{5s^2 + 7s + 2}{s^2 - 2s + 5}$ **4.3-11** For a system with transfer function

H(s) = \frac{2s+3}{s^2+2s+5}

$$- (a) Find the (zero-state) response for inputs *x*1(*t*) = 10*u*(*t*) and *x*2(*t*) = *u*(*t* −5). - (b) For this system write the differential equation relating the output *y*(*t*) to the input *x*(*t*), assuming that the systems are controllable and observable. - **4.3-12** For a system with transfer function$$

H(s) = \frac{s}{s^2 + 9}

$$- (a) Find the (zero-state) response if the input *^x*(*t*) ⁼ (1−*e*−*^t* )*u*(*t*) - (b) For this system write the differential equation relating the output *y*(*t*) to the input *x*(*t*), assuming that the systems are controllable and observable. - **4.3-13** Consider a system with transfer function$$

H(s) = \frac{s+5}{s^2 + 5s + 6}

$$Find the (zero-state) response for the following inputs: - (a) *^x*a(*t*) ⁼ *^e*−3*^t u*(*t*) - (b) *^x*b(*t*) ⁼ *^e*−4*^t u*(*t*) - (c) *^x*c(*t*) ⁼ *^e*−4(*t*−5) *u*(*t* −5) - (d) *^x*d(*t*) ⁼ *^e*−4(*t*−5) *u*(*t*) - (e) *^x*e(*t*) ⁼ *^e*−4*^t u*(*t* −5) Assuming that the system *H*(*s*) is controllable and observable, - (f) write the differential equation relating the output *y*(*t*) to the input *x*(*t*). - **4.3-14** An LTI system has a step response given by *^s*(*t*) ⁼ *^e*−*^t ^u*(*t*) ⁻ *^e*−2*^t u*(*t*). Determine the output of this system *y*(*t*) given an input *x*(*t*) = δ(*t* − π )−cos( ^√3)*u*(*t*). - **4.3-15** For an LTIC system with zero initial conditions (system initially in zero state), if an input *x*(*t*) produces an output *y*(*t*), then using the Laplace transform, show the following: - (a) The input *dx*/*dt* produces an output *dy*/*dt*. - (b) The input \$ *^t* ⁰ *x*(τ )*d*τ produces an output \$ *t* ⁰ *y*(τ )*d*τ . Hence, show that the unit step response of a system is an integral of the impulse response; that is, \$ *^t* ⁰ *h*(τ )*d*τ . - **4.3-16** Discuss asymptotic and BIBO stabilities for the systems described by the following transfer functions, assuming that the systems are controllable and observable: (a)$$

\frac{(s+5)}{s^2+3s+2}

$$(b)$$

\frac{s+5}{s^2(s+2)}

$$$$

(c) \frac{s(s+2)}{s+5}

$$(d)$$

\frac{s+5}{s(s+2)}

$$(e)$$

\frac{s+5}{s^2-2s+3}

- **4.3-17** Repeat Prob. 4.3-16 for systems described by the following differential equations. Systems may be uncontrollable and/or unobservable. - (a) (*D*² +3*D*+2)*y*(*t*) = (*D*+3)*x*(*t*) - (b) (*D*² +3*D*+2)*y*(*t*) = (*D*+1)*x*(*t*) - (c) (*D*² +*D*−2)*y*(*t*) = (*D*−1)*x*(*t*) - (d) (*D*² −3*D*+2)*y*(*t*) = (*D*−1)*x*(*t*) - **4.4-1** The circuit shown in Fig. P4.4-1 has system function given by *^H*(*s*) ⁼ ¹ ¹+*RCs*. Let *^R* ⁼ 2 and *C* = 3 and use Laplace transform techniques to solve the following. - (a) Find the output *y*(*t*) given an initial capacitor voltage of *y*(0−) = 3 and an input *x*(*t*) = *u*(*t*). - (b) Given an input *x*(*t*) = *u*(*t* − 3), determine the initial capacitor voltage *y*(0−) so that the output *y*(*t*) is 1 volt at *t* = 6 seconds. - **4.4-2** Consider the circuit shown in Fig. P4.4-2. Use Laplace transform techniques to solve the following. - (a) Determine the standard-form, constantcoefficient differential equation description of this circuit. - (b) Letting *R* = *C* = 1, determine the total response *^y*(*t*) to input *^x*(*t*) ⁼ ³*e*−*^t u*(*t*) and initial capacitor voltage of *vC*(0−) = 5. **4.4-3** Find the zero-state response *y*(*t*) of the network in Fig. P4.4-3 if the input voltage *^x*(*t*) ⁼ *te*−*^t u*(*t*). Find the transfer function relating the output *Y*(*s*) to the input *X*(*s*). From the transfer function, write the differential equation relating *y*(*t*) to *x*(*t*). ### **Figure P4.4-3** **4.4-4** The switch in the circuit of Fig. P4.4-4 is closed for a long time and then opened instantaneously at *t* = 0. Find and sketch the current *y*(*t*). #### **Figure P4.4-4** **4.4-5** Find the current *y*(*t*) for the parallel resonant circuit in Fig. P4.4-5 if the input is: (a) *x*(*t*) = *A*cos ω0*t u*(*t*) (b) *x*(*t*) = *A*sin ω0*t u*(*t*) Assume all initial conditions to be zero and, in both cases, ω² ⁰ = 1/*LC*. **Figure P4.4-5** **Figure P4.4-6** - **4.4-6** Find the loop currents *y*1(*t*) and *y*2(*t*) for *t* ≥ 0 in the circuit of Fig. P4.4-6a for the input *x*(*t*) in Fig. P4.4-6b. - **4.4-7** For the network in Fig. P4.4-7, the switch is in a closed position for a long time before *t* = 0, when it is opened instantaneously. Find *y*1(*t*) and *vs*(*t*) for *t* ≥ 0. #### **Figure P4.4-7** - **4.4-8** Find the output voltage *v*0(*t*) for *t* ≥ 0 for the circuit in Fig. P4.4-8, if the input *x*(*t*) = 100*u*(*t*). The system is in the zero state initially. - **4.4-9** Find the output voltage *y*(*t*) for the network in Fig. P4.4-9 for the initial conditions *iL*(0) = 1 A and *vC*(0) = 3 V. - **4.4-10** For the network in Fig. P4.4-10, the switch is in position *a* for a long time and then is moved to position *b* instantaneously at *t* = 0. Determine the current *y*(*t*) for *t* > 0. **4.4-11** Consider the circuit of Fig. P4.4-11. - (a) Using transform-domain techniques, determine the system's standard-form transfer function *H*(*s*). - (b) Using transform-domain techniques and letting *R* = *L* = 1, determine the circuit's zero-state response *y*zsr(*t*) to the input *x*(*t*) = *e*−2*^t u*(*t* −1). **Figure P4.4-10** (c) Using transform-domain techniques and letting *R* = 2*L* = 1, determine the circuit's zero-state response *y*zsr(*t*) to the input *x*(*t*) = *e*−2*^t u*(*t* −1). **4.4-12** Show that the transfer function that relates the output voltage *y*(*t*) to the input voltage *x*(*t*) for the op-amp circuit in Fig. P4.4-12a is given by

H(s) = \frac{Ka}{s+a} \quad \text{where}

$$$$

K = 1 + \frac{R_b}{R_a} \quad \text{and} \quad a = \frac{1}{RC}

$$and that the transfer function for the circuit in Fig. P4.4-12b is given by$$

H(s) = \frac{Ks}{s+a}

**4.4-13** For the second-order op-amp circuit in Fig. P4.4-13, show that the transfer function *H*(*s*) relating the output voltage *y*(*t*) to the input ### **Figure P4.4-14** voltage *x*(*t*) is given by

H(s) = \frac{-s}{s^2 + 8s + 12}

$$- **4.4-14** Consider the op-amp circuit of Fig. P4.4-14. - (a) Determine the standard-form transfer function *H*(*s*) of this system. - (b) Determine the standard-form constant coefficient linear differential equation description of this circuit. - (c) Using transform-domain techniques, determine the circuit's zero-state response *y*zsr(*t*) to the input *^x*(*t*) ⁼ *^e*2*^t u*(*t* +1). - (d) Using transform-domain techniques, determine the circuit's zero-input response *y*zir(*t*) if the *t* = 0⁻ capacitor voltage (first op-amp output voltage) is 3 volts. - **4.4-15** We desire the op-amp circuit of Fig. P4.4-15 to behave as *y*˙(*t*)−1.5*y*(*t*) = −3*x*˙(*t*)+0.75*x*(*t*). - (a) Determine resistors *R*1, *R*2, and *R*³ so that the circuit's input–output behavior follows the desired differential equation of *y*˙(*t*) − 1.5*y*(*t*) = −3*x*˙(*t*)+0.75*x*(*t*). - (b) Using transform-domain techniques, determine the circuit's zero-input response *y*zir(*t*) if the *t* = 0 capacitor voltage (first op-amp output voltage) is 2 volts. - (c) Using transform-domain techniques, determine the impulse response *h*(*t*) of this circuit. - (d) Determine the circuit's zero-state response *y*zsr(*t*) to the input *x*(*t*) = *u*(*t* −2). - **4.4-16** We desire the op-amp circuit of Fig. P4.4-16 to behave as \$ \$ *^y*(*t*)⁺ ² 5 \$ *^y*(*t*)⁺ ¹ ⁵ *^y*(*t*) ⁼ \$ \$ *^x*(*t*)⁻ \$ *x*(*t*) - (a) Determine the resistors *R*1, *R*2, and *R*³ to produce the desired behavior. - (b) Using transform-domain techniques, determine the circuit's zero-input response *y*zir(*t*) if the *t* = 0 capacitor voltages (first two op-amp outputs) are each 1 volt. **4.4-17** (a) Using the initial and final value theorems, find the initial and final value of the zero-state response of a system with the transfer function$$

H(s) = \frac{6s^2 + 3s + 10}{2s^2 + 6s + 5}

$$and input *x*(*t*) = *u*(*t*). (b) Repeat part (a) for the input *^x*(*t*) ⁼ *^e*−*^t u*(*t*). (c) Find y(0⁺) and y($$

\infty

) if $Y(s) = \frac{s^2 + 5s + 6}{s^2 + 3s + 2}$ . - (d) Find *y*(0+) and *y*(∞) if *Y*(*s*) = *s*³ +4*s*² +10*s*+7 *^s*² ⁺2*^s* ⁺³ . - **4.5-1** Consider two LTIC systems. The first has transfer function *^H*1(*s*) ⁼ ²*^s ^s*+¹ , and the second has transfer function *^H*2(*s*) ⁼ ¹ *se*3(*s*−1) . **Figure P4.5-2** - (a) Determine the overall impulse response *h*s(*t*) if the two systems are connected in series. - (b) Determine the overall impulse response *h*p(*t*) if the two systems are connected in parallel. - **4.5-2** Figure P4.5-2a shows two resistive ladder segments. The transfer function of each segment (ratio of output to input voltage) is 1/2. Figure P4.5-2b shows these two segments connected in cascade. - (a) Is the transfer function (ratio of output to input voltage) of this cascaded network (1/2)(1/2) = 1/4? - (b) If your answer is affirmative, verify the answer by direct computation of the transfer function. Does this computation confirm the earlier value 1/4? If not, why? - (c) Repeat the problem with *R*³ = *R*⁴ = 20 k. Does this result suggest the answer to the problem in part (b)? - **4.5-3** In communication channels, transmitted signal is propagated simultaneously by several paths of varying lengths. This causes the signal to reach the destination with varying time delays and varying gains. Such a system generally distorts the received signal. For error-free communication, it is necessary to undo this distortion as For simplicity, let us assume that a signal is propagated by two paths whose time delays differ by τ seconds. The channel over the intended path has a delay of *T* seconds and unity gain. The signal over the unintended path has a delay of *T* + τ seconds and gain *a*. Such a channel can be modeled, as shown in Fig. P4.5-3. Find the inverse system transfer function to correct the delay distortion and show that the inverse system can be realized by a feedback system. The inverse system should be causal to be realizable. [*Hint:* We want to correct only the distortion caused by the relative delay τ seconds. For distortionless transmission, the signal may be delayed. What is important is to maintain the shape of *x*(*t*). Thus, a received signal of the form *c x*(*t* −*T*) is considered to be distortionless.] **4.5-4** Discuss BIBO stability of the feedback systems depicted in Fig. P4.5-4. For the system in **Figure P4.5-4** Fig. P4.5-4b, consider three cases: **(a)** *K* = 10, **(b)** *K* = 50, and **(c)** *K* = 48. **4.6-1** Realize

H(s) = \frac{s(s+2)}{(s+1)(s+3)(s+4)}

$$by canonic direct, series, and parallel forms. - **4.6-2** Realize the transfer function in Prob. 4.6-1 by using the transposed form of the realizations found in Prob. 4.6-1. - **4.6-3** Repeat Prob. 4.6-1 for (a) *^H*(*s*) ⁼ ³*s*(*^s* ⁺2) (*s* +1)(*s*² +2*s*+2) (b) *^H*(*s*) ⁼ ²*^s* ⁻⁴ (*s* +2)(*s*² +4) - **4.6-4** Realize the transfer functions in Prob. 4.6-3 by using the transposed form of the realizations found in Prob. 4.6-3. - **4.6-5** Repeat Prob. 4.6-1 for$$

H(s) = \frac{2s+3}{5s(s+2)^2(s+3)}

$$- **4.6-6** Realize the transfer function in Prob. 4.6-5 by using the transposed form of the realizations found in Prob. 4.6-5. - **4.6-7** Repeat Prob. 4.6-1 for$$

H(s) = \frac{s(s+1)(s+2)}{(s+5)(s+6)(s+8)}

$$- **4.6-8** Realize the transfer function in Prob. 4.6-7 by using the transposed form of the realizations found in Prob. 4.6-7. - **4.6-9** Repeat Prob. 4.6-1 for$$

H(s) = \frac{s^3}{(s+1)^2(s+2)(s+3)}

$$- **4.6-10** Realize the transfer function in Prob. 4.6-9 by using the transposed form of the realizations found in Prob. 4.6-9. - **4.6-11** Repeat Prob. 4.6-1 for$$

H(s) = \frac{s^3}{(s+1)(s^2+4s+13)}

$$**4.6-12** Realize the transfer function in Prob. 4.6-11 by using the transposed form of the realizations found in Prob. 4.6-11. - **4.6-13** Draw a TDFII block realization of a causal LTIC system with transfer function *^H*(*s*) ⁼ (*s*−2*j*)(*s*+2*j*) (*s*−*j*)(*s*+*j*)(*s*+2) . Give two reasons why TDFII tends to be a good structure. - **4.6-14** Consider a causal LTIC system with transfer function *^H*(*s*) ⁼ (*s*−2*j*)(*s*+2*j*)(*s*−3*j*)(*s*+3*j*) ⁹(*s*+1)(*s*+2)(*s*+1−*j*)(*s*+1+*j*) . - (a) Realize *H*(*s*) using a single fourth-order real TDFII structure. Is this block realization unique? Explain. - (b) Realize *H*(*s*) using a cascade of secondorder real DFII structures. Is this block realization unique? Explain. - (c) Realize *H*(*s*) using a parallel connection of second-order real DFI structures. Is this block realization unique? Explain. - **4.6-15** In this problem we show how a pair of complex conjugate poles may be realized by using a cascade of two first-order transfer functions and feedback. Show that the transfer functions of the block diagrams in Figs. P4.6-15a and P4.6-15b are: (a)$$

H_a(s) = \frac{1}{(s+a)^2 + b^2}

$$=$$

\frac{1}{s^2 + 2as + (a^2 + b^2)}

$$(b)$$

Hb(s) = \frac{s+a}{(s+a)^2 + b^2}

$$=$$

\frac{s+a}{s^2 + 2as + (a^2 + b^2)}

$$Hence, show that the transfer function of the block diagram in Fig. P4.6-15c is (c)$$

H_c(s) = \frac{As + B}{(s+a)^2 + b^2}

$$=$$

\frac{As + B}{s^2 + 2as + (a^2 + b^2)}

$$**4.6-16** Show op-amp realizations of the following transfer functions: (a)$$

\frac{-10}{s+5}

\n(b) $\frac{10}{s+5}$ \n(c) $\frac{s+2}{s+5}$ (c) *b*2 **4.6-17** Show two different op-amp circuit realizations of the transfer function

H(s) = \frac{s+2}{s+5} = 1 - \frac{3}{s+5}

$$**4.6-18** Show an op-amp canonic direct realization of the transfer function$$

H(s) = \frac{3s + 7}{s^2 + 4s + 10}

$$**4.6-19** Show an op-amp canonic direct realization of the transfer function$$

H(s) = \frac{s^2 + 5s + 2}{s^2 + 4s + 13}

$$**4.6-20** Consider a system described by a constantcoefficient linear differential equation as *^d dt y*(*t*) ⁺ ²*y*(*t*) ⁼ *^x*(*t*) ⁻ ³ *^d dt x*(*t*). Draw an op-amp realization of this system if resistors and inductors are available but not capacitors. Would using inductors rather than capacitors in this circuit pose any problem? Explain. - **4.7-1** Feedback can be used to increase (or decrease) the system bandwidth. Consider the system in Fig. P4.7-1a with transfer function *G*(*s*) = ω*c*/(*s* +ω*c*). - (a) Show that the 3 dB bandwidth of this system is ω*^c* and the dc gain is unity; that is, |*H*(*j*0)| = 1. - (b) To increase the bandwidth of this system, we use negative feedback with *H*(*s*) = 9, as depicted in Fig. P4.7-1b. Show that the 3 dB bandwidth of this system is 10ω*c*. What is the dc gain? - (c) To decrease the bandwidth of this system, we use positive feedback with *H*(*s*) = −0.9, as illustrated in Fig. P4.7-1c. Show that the 3 dB bandwidth of this system is ω*c*/10. What is the dc gain? - (d) The system gain at dc times its 3 dB bandwidth is the *gain-bandwidth product* of a system. Show that this product is the same for all the three systems in Fig. P4.7-1. This result shows that if we increase the bandwidth, the gain decreases and vice versa. - **4.8-1** Suppose an engineer builds a controllable, observable LTIC system with transfer function *^H*(*s*) ⁼ *^s*2+⁴ ²*s*2+4*s*+⁴ . - (a) By direct calculation, compute the magnitude response at frequencies ω = 0, 1, 2, 3, 5, 10, and ∞. Use these calculations to roughly sketch the magnitude response over 0 ≤ ω ≤ 10. - (b) To test the system, the engineer connects a signal generator to the system in hopes to measure the magnitude response using a standard oscilloscope. What type of signal should the engineer input into the system to make the measurements? How should the engineer make the measurements? Provide sufficient detail to fully justify your answers. - (c) Suppose the engineer accidentally constructs the system *^H*−1(*s*)⁼ ¹ *^H*(*s*) ⁼ ²*s*2+4*s*+⁴ *^s*2+⁴ . What impact will this mistake have on his tests? - **4.8-2** For an LTIC system described by the transfer function$$

H(s) = \frac{s+2}{s^2 + 5s + 4}

$$find the response to the following everlasting sinusoidal inputs: - (a) 5 cos(2*t* +30◦) - (b) 10 sin(2*t* +45◦) - (c) 10 cos(3*t* +40◦) Observe that these are everlasting sinusoids. **4.8-3** For an LTIC system described by the transfer function$$

H(s) = \frac{s+3}{(s+2)^2}

$$find the steady-state system response to the following inputs: - (a) 10*u*(*t*) - (b) cos(2*t* +60◦)*u*(*t*) - (c) sin(3*t* −45◦)*u*(*t*) - (d) *ej*3*^t u*(*t*) - **4.8-4** For an allpass filter specified by the transfer function$$

H(s) = \frac{-(s-10)}{s+10}

find the system response to the following (everlasting) inputs: - (a) *ej*ω*^t* - (b) cos(ω*t* +θ ) - (c) cos *t* - (d) sin 2*t* - (e) cos 10*t* - (f) cos 100*t* Comment on the filter response. - **4.8-5** The pole-zero plot of a second-order system *H*(*s*) is shown in Fig. P4.8-5. The dc response of this system is minus 1, *H*(*j*0) = −1. - (a) Letting *H*(*s*) = *k*(*s*² +*b*1*s*+*b*2)/(*s*² +*a*1*s*+ *a*2), determine the constants *k*, *b*1, *b*2, *a*1, and *a*2. - (b) What is the output *y*(*t*) of this system in response to the input *x*(*t*) = 4 + cos(*t*/2 + π/3)? ### **Figure P4.8-5** - **4.8-6** Consider a CT system described by (*D*+1)(*D*+ 2){*y*(*t*)} = *x*(*t* − 1). Notice that this differential equation is in terms of *x*(*t* −1), not *x*(*t*)! - (a) Determine the output *y*(*t*) given input *x*(*t*) = 1. - (b) Determine the output *y*(*t*) given input *x*(*t*) = cos(*t*). - **4.8-7** An LTIC system has transfer function *^H*(*s*) ⁼ ⁴*^s ^s*2+2*s*+³⁷ ⁼ ⁴*^s* (*s*+1+6*j*)(*s*+1−6*j*). Determine the steady-state output in response to input *^x*(*t*) ⁼ ¹ ³ *ej*(6*t*+π/3) *u*(6*t* +π/3). - **4.9-1** Suppose a real first-order lowpass system *H*(*s*) has unity gain in the passband, one finite pole at *s* = −2, and one finite zero at an unspecified location. - (a) Determine the location of the system zero so that the filter achieves 40 dB of stopband attenuation. Sketch the corresponding straight-line Bode approximation of the system magnitude response. - (b) Determine the location of the system zero so that the filter achieves 30 dB of stopband attenuation. Sketch the corresponding straight-line Bode approximation of the system magnitude response. - **4.9-2** Repeat Prob. 4.9-1 for a highpass rather than a lowpass system. - **4.9-3** Repeat Prob. 4.9-1 for a second-order system that has a pair of repeated poles and a pair of repeated zeros. - **4.9-4** Sketch Bode plots for the following transfer functions: (a)

\frac{s(s+100)}{(s+2)(s+20)}

$$(b)$$

\frac{(s+10)(s+20)}{s^2(s+100)}

$$(c)$$

\frac{(s+10)(s+200)}{(s+20)^2(s+1000)}

$$$$

4.9-5 \quad \text{Repeat Prob. } 4.9-4 \text{ for}

$$(a)$$

\frac{s^2}{(s+1)(s^2+4s+16)}

$$\n(b)$$

\frac{s}{(s+1)(s^2+14.14s+100)}

$$\n(c)$$

\frac{(s+10)}{s(s^2+14.14s+100)}

$$- **4.9-6** Using the lowest order possible, determine a system function *H*(*s*) with real-valued roots that matches the frequency response in Fig. P4.9-6. Verify your answer with MATLAB. - **4.9-7** A graduate student recently implemented an analog phase lock loop (PLL) as part of his thesis. His PLL consists of four basic components: a phase/frequency detector, a charge pump, a loop filter, and a voltage-controlled oscillator. This problem considers only the loop filter, which is shown in Fig. P4.9-7a. The loop filter input is the current *x*(*t*), and the output is the voltage *y*(*t*). - (a) Derive the loop filter's transfer function *H*(*s*). Express *H*(*s*) in standard form. - (b) Figure P4.9-7b provides four possible frequency response plots, labeled A through D. Each log-log plot is drawn to the same scale, and line slopes are either 20 dB/decade, 0 dB/decade, or −20 dB/decade. Clearly identify which plot(s), if any, could represent the loop filter. - (c) Holding the other components constant, what is the general effect of increasing the resistance *R* on the magnitude response for low-frequency inputs? **Figure P4.9-6** (d) Holding the other components constant, what is the general effect of increasing the resistance *R* on the magnitude response for high-frequency inputs? **Figure P4.9-7** - **4.10-1** A causal LTIC system *^H*(*s*) ⁼ ²(*s*−4*j*)(*s*+4*j*) (*s*+1+2*j*)(*s*+1−2*j*) has input *x*(*t*) = −1 + 2 cos(2*t*) − 3 sin(4*t* + π/3) + 4 cos(10*t*). Below, perform accurate calculations at ω = 0, ±2, ±4, and ±10. - (a) Using the graphical method of Sec. 4.10-1, accurately sketch the magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10. - (b) Using the graphical method of Sec. 4.10-1, accurately sketch the phase response *H*(*j*ω) over −10 ≤ ω ≤ 10. - (c) Approximate the system output *y*(*t*) in response to the input *x*(*t*). - **4.10-2** The pole-zero plot of a second-order system *H*(*s*) is shown in Fig. P4.10-2. The dc response of this system is minus 2, *H*(*j*0) = −2. - (a) Letting *^H*(*s*) ⁼ *^k ^s*2+*b*1*s*+*b*² *^s*2+*a*1*s*+*a*² , determine the constants *k*, *b*1, *b*2, *a*1, and *a*2. - (b) Using the graphical method of Sec. 4.10-1, hand-sketch the magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10. Verify your sketch with MATLAB. - (c) Using the graphical method of Sec. 4.10-1, hand-sketch the phase response *H*(*j*ω) over −10 ≤ ω ≤ 10. Verify your sketch with MATLAB. - (d) What is the output *y*(*t*) in response to input *x*(*t*) = −3+cos(3*t*+π/3)−sin(4*t*−π/8)? **Figure P4.10-2** **4.10-3** Using the graphical method of Sec. 4.10-1, draw a rough sketch of the amplitude and phase responses of an LTIC system described by the transfer function$$

H(s) = \frac{s^2 - 2s + 50}{s^2 + 2s + 50}

$$=$$

\frac{(s - 1 - j7)(s - 1 + j7)}{(s + 1 - j7)(s + 1 + j7)}

$$What kind of filter is this? **4.10-4** Using the graphical method of Sec. 4.10-1, draw a rough sketch of the amplitude and phase responses of LTIC systems whose pole-zero plots are shown in Fig. P4.10-4. **Figure P4.10-4** - **4.10-5** A causal LTIC system *^H*(*s*) ⁼ (*s*−3*j*)(*s*+3*j*) 3(*s*+2+*j*)(*s*+2−*j*) has input *x*(*t*) = cos(*t*) + sin(3*t* + π/3) + cos(100*t*). - (a) Using the graphical method of Sec. 4.10-1, sketch the magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10. - (b) Determine the system output *y*(*t*) in response to the input *x*(*t*). - (c) Suppose we create a second causal system with transfer function *H*2(*s*) = *H*(−*s*). Sketch this system's pole/zero plot. What is the response *y*2(*t*) of system *H*2(*s*) to input *x*(*t*)? - **4.10-6** Design a second-order bandpass filter with center frequency ω = 10. The gain should be zero at ω = 0 and at ω = ∞. Select poles at −*a* ± *j*10. Leave your answer in terms of *a*. Explain the influence of *a* on the frequency response. - **4.10-7** The LTIC system described by *H*(*s*) = (*s*−1)/(*s* +1) has unity magnitude response |*H*(*j*ω)| = 1. Positive Pat claims that the output *y*(*t*) of this system is equal the input *x*(*t*), since the system is allpass. Cynical Cynthia doesn't think so. "This is *signals and systems* class," she complains. "It *has* to be more complicated!" Who is correct, Pat or Cynthia? Justify your answer. - **4.10-8** Two students, Amy and Jeff, disagree about an analog system function given by *H*1(*s*) = *s*. Sensible Jeff claims the system has a zero at *s* = 0. Rebellious Amy, however, notes that the system function can be rewritten as *H*1(*s*) = 1/*s*−¹ and claims that this implies a system pole at *s* = ∞. Who is correct? Why? What are the poles and zeros of the system *H*2(*s*) = 1/*s*? - **4.10-9** A rational transfer function *H*(*s*) is often used to represent an analog filter. Why must *H*(*s*) be strictly proper for lowpass and bandpass filters? Why must *H*(*s*) be proper for highpass and bandstop filters? - **4.10-10** For a given filter order *N*, why is the stopband attenuation rate of an all-pole lowpass filter better than filters with finite zeros? - **4.10-11** Is it possible, with real coefficients ([*k*,*b*1,*b*2,*a*1,*a*2] ∈ *R*), for a system$$

H(s) = k \frac{s^2 + b_1 s + b_2}{s^2 + a_1 s + a_2}

$$to function as a lowpass filter? Explain your answer. - **4.10-12** Nick recently built a simple second-order Butterworth lowpass filter for his home stereo. Although the system performs fairly well, Nick is an overachiever and hopes to improve the system performance. Unfortunately, Nick is lazy and doesn't want to design another filter. Thinking "Twice the filtering gives twice the performance," he suggests filtering the audio signal not once but twice with a cascade of two identical filters. His overworked, underpaid signals professor is skeptical and states, "If you are using *identical* filters, it makes no difference whether you filter once or twice!" Who is correct? Why? - **4.10-13** An LTIC system impulse response is given by *h*(*t*) = *u*(*t*)−*u*(*t* −1). - (a) Determine the transfer function *H*(*s*). Using *H*(*s*), determine and plot the magnitude response |*H*(*j*ω)|. Which type of filter most accurately describes the behavior of this system: lowpass, highpass, bandpass, or bandstop? - (b) What are the poles and zeros of *H*(*s*)? Explain your answer. - (c) Can you determine the impulse response of the inverse system? If so, provide it. If not, suggest a method that could be used to approximate the impulse response of the inverse system. - **4.10-14** An ideal lowpass filter *H*LP(*s*) has magnitude response that is unity for low frequencies and zero for high frequencies. An ideal highpass filter *H*HP(*s*) has an opposite magnitude response: zero for low frequencies and unity for high frequencies. A student suggests a possible lowpass-to-highpass filter transformation: *H*HP(*s*) = 1 − *H*LP(*s*). In general, will this transformation work? Explain your answer. - **4.10-15** An LTIC system has a rational transfer function *H*(*s*). When appropriate, assume that all initial conditions are zero. - (a) Is is possible for this system to output *y*(*t*) = sin(100π*t*)*u*(*t*) in response to an input *x*(*t*) = cos(100π*t*)*u*(*t*)? Explain. - (b) Is is possible for this system to output *y*(*t*) = sin(100π*t*)*u*(*t*) in response to an input *x*(*t*) = sin(50π*t*)*u*(*t*)? Explain. - (c) Is is possible for this system to output *y*(*t*) = sin(100π*t*) in response to an input *x*(*t*) = cos(100π*t*)? Explain. - (d) Is is possible for this system to output *y*(*t*) = sin(100π*t*) in response to an input *x*(*t*) = sin(50π*t*)? Explain. - **4.11-1** Find the ROC, if it exists, of the (bilateral) Laplace transform of the following signals: - (a) *etu*(*t*) (b) *e*−*tu*(*t*) (c)$$

\frac{1}{1+t^2}

$$(d)$$

\frac{1}{1+e^t}

(e) $e^{-kt^2}$ - **4.11-2** Using the definition and direct integration, find the (bilateral) Laplace transform and the corresponding region of convergence for the following signals. If the Laplace transform does not exist, carefully explain why. - (a) *^x*a(*t*) ⁼ *^e*(−1−*j*)*^t u*(1−*t*) - (b) *x*b(*t*) = *j* (*t*+1) *u*(−*t* −1) - (c) *^x*c(*t*) ⁼ *ej*π/3*u*(2−*t*) ⁺*j*δ(*^t* ⁻5) - (d) *x*d(*t*) = 1+1 = 2 - (e) *^x*e(*t*) ⁼ ³*u*(−*t*) ⁺*e*−2*^t* [*u*(*t*)−*u*(*t* −10)] - (f) *^x*f(*t*) ⁼ *et*−2*u*(1−*t*) ⁺*e*−2*^t u*(*t* +1) - **4.11-3** Determine the bilateral Laplace transform *X*(*s*) of the signal

x(t) = \left[e^t u(-t)\right] * \left[t\cos(2t)u(t)\right]

$$**4.11-4** A signal has bilateral Laplace transform *^X*(*s*) ⁼ (*s*+1) (*s*−2)(*s*−3) but unknown region of convergence. What ROC results in the smallest maximum amplitude of *x*(*t*)? Justify your answer. - **4.11-5** Find the (bilateral) Laplace transform and the corresponding region of convergence for the following signals: - (a) *e*−|*t*^| - (b) *e*−|*t*^| cos *t* (c)$$

e^t u(t) + e^{2t} u(-t)

(d) $e^{-tu(t)}$

\begin{array}{cc}\n(e) & e^{-tu(-t)} \ (e) & e^{-tu(-t)}\n\end{array}

$$- (f) cosω0*t u*(*t*) ⁺*et ^u*(−*t*) - **4.11-6** Find the inverse (bilateral) Laplace transforms of the following functions: (a)$$

\frac{2s+5}{(s+2)(s+3)}

$$\n$$

-3 < \sigma < -2

$$\n(b)$$

\frac{2s-5}{(s-2)(s-3)}

$$\n$$

2 < \sigma < 3

$$\n(c)$$

\frac{2s+3}{(s+1)(s+2)}

$$\n$$

\sigma > -1

$$\n(d)$$

\frac{2s+3}{(s+1)(s+2)}

$$\n$$

\sigma < -2

$$(d)$$

\frac{2s+6}{(s+1)(s+2)} \quad \sigma < -2

$$(e)$$

\frac{3s^2 - 2s - 17}{(s+1)(s+3)(s-5)} \quad -1 < \sigma < 5

$$**4.11-7** Find$$

\mathcal{L}^{-1}\left[\frac{2s^2 - 2s - 6}{(s+1)(s-1)(s+2)}\right]

$$if the ROC is - (a) Re *s* > 1 - (b) Re *s* < −2 - (c) −1 < Re*s* < 1 - (d) −2 < Re*s* < −1 - **4.11-8** For a causal LTIC system having a transfer function *H*(*s*) = 1/(*s*+1), find the output *y*(*t*) if the input *x*(*t*) is given by (a) *e*−|*t*|/² (b) *et ^u*(*t*)+*e*2*^t u*(−*t*) (c) *^e*−*t*/2*u*(*t*) ⁺*e*−*t*/4*u*(−*t*) (d) *e*2*^t ^u*(*t*)+*et u*(−*t*) (e) *^e*−*t*/4*u*(*t*) ⁺*e*−*t*/2*u*(−*t*) - (f) *e*−3*^t ^u*(*t*)+*e*−2*^t u*(−*t*) - **4.11-9** The autocorrelation function *rxx*(*t*) of a signal *x*(*t*) is given by$$

r_{xx}(t) = \int_{-\infty}^{\infty} x(\tau) x(\tau + t) d\tau

$$Derive an expression for *Rxx*(*s*) = *L*(*rxx*(*t*)) in terms of *X*(*s*), where *X*(*s*) = *L*(*x*(*t*)). **4.11-10** Determine the inverse Laplace transform of$$

X(s) = \frac{2}{s} + \frac{s}{2}

$$given that the region of convergence is σ < 0. **4.11-11** An absolutely integrable signal *x*(*t*) has a pole at *s* = π. It is possible that other poles may be present. Recall that an absolutely integrable signal satisfies$$

\int_{-\infty}^{\infty} |x(t)| dt < \infty

$$- (a) Can *x*(*t*) be left-sided? Explain. - (b) Can *x*(*t*) be right-sided? Explain. - (c) Can *x*(*t*) be two-sided? Explain. - (d) Can *x*(*t*) be of finite duration? Explain. - **4.11-12** Using ROC σ < 0, determine the inverse Laplace transform of *^X*(*s*) ⁼ *^s ^d ds e*−2*^s s* . [*Hint:* Use Laplace transform properties to avoid tedious calculus.] - **4.11-13** With the assistance of Laplace transform properties, determine the inverse bilateral Laplace transform *x*(*t*) of signal$$

X(s) = \frac{2}{e^s} + \frac{1}{s} \left[ e^s \frac{4}{\frac{s}{3} + 2} \right]

$$where the ROC is −6 < Re{*s*} < 0. **4.11-14** With the assistance of Laplace transform properties, determine the inverse bilateral Laplace transform *x*(*t*) of signal$$

X(s) = \frac{d^7}{ds^7} \left[ \frac{e^{-4s}}{(s+2)(s+3)} \right]

$$where the ROC is −3 < Re{*s*} < −2. **4.11-15** Using the definition, compute the bilateral Laplace transform, including the region of convergence (ROC), of the following complexvalued functions: (a)$$

x_1(t) = (j + e^{jt})u(t)

$$- (b) *x*2(*t*) = *j* cosh(*t*)*u*(−*t*) - (c) *^x*3(*t*) ⁼ *ej*( ^π 4 ) *u*(−*t* +1)+*j*δ(*t* −5) - (d) *x*4(*t*) = *j t u*(−*t*) +δ(*t* −π ) **4.11-16** A bounded-amplitude signal *x*(*t*) has bilateral Laplace transform *X*(*s*) given by$$

X(s) = \frac{s2^s}{(s-1)(s+1)}

$$- (a) Determine the corresponding region of convergence. - (b) Determine the time-domain signal *x*(*t*). - **4.12-1** Express the polynomial *C*20(*x*) in standard form. That is, determine the coefficients - **4.12-2** Consider an LTIC system with$$

H(s) = \frac{1}{s^3 + 4s^2 + 8s + 8} = \frac{1}{(s^2 + 2s + 4)(s + 2)}

$$=$$

\frac{1}{(s - 2e^{j2\pi/3})(s - 2e^{-j2\pi/3})(s + 2)}.

- (a) Write MATLAB code that accurately plots the system magnitude response |*H*(*j*ω)| over −10 ≤ ω ≤ 10. - (b) Write MATLAB code that accurately plots the system phase response over −10 ≤ ω ≤ 10. - (c) Determine the max value *y*max of output *y*(*t*) in response to input *x*(*t*) = 2−sin(2*t*+π/3). - (d) Draw a parallel representation of this system using real DFI structures of order 2 or less. [*Hint:* Use MATLAB to perform a partial fraction expansion of *H*(*s*).] - **4.12-3** Consider the op-amp circuit of Fig. P4.12-3. Further, let *RC* = 1. - (a) From Fig. P4.12-3, determine the (simplified, standard form, rational) transfer function *H*(*s*). - (b) Use MATLAB to accurately plot |*H*(*j*ω)| over −10 ≤ ω ≤ 10. - (c) Determine the output of this system in response to *x*(*t*) = cos(10*t*)−1. - (d) The circuit of Fig. P4.12-3 contains two capacitors. Suppose one capacitor must be a 25% tolerance part, while the other must be a 10% tolerance part. If the goal is to preserve the original magnitude response, should you use the 25% tolerance capacitor with the first op-amp or the second op-amp? Justify your answer with appropriate MAT-LAB simulations. **Figure P4.12-3** - **4.12-4** Design an order-12 Butterworth lowpass filter with a cutoff frequency of ω*^c* = 2π5000 by completing the following. - (a) Locate and plot the filter's poles and zeros in the complex plane. Plot the corresponding magnitude response |*H*LP(*j*ω)| to verify proper design. - (b) Setting all resistor values to 100,000, determine the capacitor values to implement the filter using a cascade of six second-order Sallen–Key circuit sections. The form of a Sallen–Key stage is shown in Fig. P4.12-4. On a single plot, plot the magnitude response of each section as well as the overall magnitude response. Identify the poles that correspond to each section's magnitude response curve. Are the capacitor values realistic? - **4.12-5** Rather than a Butterworth filter, repeat Prob. 4.12-4 for a Chebyshev LPF with *R* = 3 dB of passband ripple. Since each Sallen–Key stage is constrained to have unity gain at dc, an overall gain error of 1/ √ 1+ ² is acceptable. - **4.12-6** An analog lowpass filter with cutoff frequency ω*^c* can be transformed into a highpass filter with cutoff frequency ω*^c* by using an *RC*–*CR* transformation rule: each resistor *Ri* is replaced by a capacitor *C ⁱ* = 1/*Ri*ω*^c* and each capacitor *Ci* is replaced by a resistor *R ⁱ* = 1/*Ci*ω*c*. Use this rule to design an order-8 Butterworth highpass filter with ω*^c* = 2π4000 by completing the following. - (a) Design an order-8 Butterworth lowpass filter with ω*^c* = 2π4000 by using four second-order Sallen–Key circuit stages, the form of which is shown in Fig. P4.12-4. Give resistor and capacitor values for each stage. Choose the resistors so that the *RC*–*CR* transformation will result in 1 nF capacitors. At this point, are the component values realistic? - (b) Draw an *RC*–*CR* transformed Sallen–Key circuit stage. Determine the transfer function *H*(*s*) of the transformed stage in terms of the variables *R* ¹, *R* ², *C* ¹, and *C* 2. - (c) Transform the LPF designed in part (a) by using an *RC*–*CR* transformation. Give the resistor and capacitor values for each stage. Are the component values realistic? Using *H*(*s*) derived in part (b), plot the magnitude response of each section as well as the overall magnitude response. Does the overall response look like a highpass Butterworth filter? Plot the HPF system poles and zeros in the complex *s* plane. How do these locations compare with those of the Butterworth LPF? - **4.12-7** Repeat Prob. 4.12-6, using ω*^c* = 2π1500 and an order-16 filter. That is, eight second-order stages need to be designed. - **4.12-8** Rather than a Butterworth filter, repeat Prob. 4.12-6 for a Chebyshev LPF with *R* = 3 dB of passband ripple. Since each transformed Sallen–Key stage is constrained to have unity gain at ω = ∞, an overall gain error of 1/ √ 1+ ² is acceptable. - **4.12-9** The MATLAB signal-processing toolbox function butter helps design analog Butterworth filters. Use MATLAB help to learn how butter works. For each of the following cases, design the filter, plot the filter's poles and zeros in the complex *s* plane, and plot the decibel magnitude response 20log10 |*H*(*j*ω)|: - (a) Design a sixth-order analog lowpass filter with ω*^c* = 2π3500. - (b) Design a sixth-order analog highpass filter with ω*^c* = 2π3500. - (c) Design a sixth-order analog bandpass filter with a passband between 2 and 4 kHz. - (d) Design a sixth-order analog bandstop filter with a stopband between 2 and 4 kHz. - **4.12-10** The MATLAB signal-processing toolbox function cheby1 helps design analog Chebyshev type I filters. A Chebyshev type I filter has a passband ripple and a smooth stopband. Setting the passband ripple to *Rp* = 3 dB, repeat Prob. 4.12-9 using the cheby1 command. With all other parameters held constant, what is the general effect of reducing *Rp*, the allowable passband ripple? - **4.12-11** The MATLAB signal-processing toolbox function cheby2 helps design analog Chebyshev type II filters. A Chebyshev type II filter has a smooth passband and ripple in the stopband. Setting the stopband ripple *Rs* = 20 dB down, repeat Prob. 4.12-9 using the cheby2 command. With all other parameters held constant, what is the general effect of increasing *Rs*, the minimum stopband attenuation? - **4.12-12** The MATLAB signal-processing toolbox function ellip helps design analog elliptic filters. An elliptic filter has ripple in both the passband and the stopband. Setting the passband ripple to *Rp* = 3 dB and the stopband ripple *Rs* = 20 dB down, repeat Prob. 4.12-9 using the ellip command. - **4.12-13** Using the definition *CN*(*x*)=cosh(*^N* cosh−1(*x*)), prove the recursive relation *CN*(*x*) = 2*xCN*−1(*x*) −*CN*−2(*x*). - **4.12-14** Prove that the poles of a Chebyshev filter, which are located at *pk* = ω*^c* sinh(ξ )sin(φ*k*) + *j*ω*^c* cosh(ξ ) cos(φ*k*), lie on an ellipse. [*Hint:* The equation of an ellipse in the *x*–*y* plane is (*x*/*a*) 2+ (*y*/*b*) ² ⁼ 1, where constants *^a* and *^b* define the major and minor axes of the ellipse.] # **[DISCRETE-TIME](#page-11-0) SYSTEM ANALYSIS USING THE** *z***-TRANSFORM** The counterpart of the Laplace transform for discrete-time systems is the *z*-transform. The Laplace transform converts integro-differential equations into algebraic equations. In the same way, the *z*-transforms changes difference equations into algebraic equations, thereby simplifying the analysis of discrete-time systems. The *z*-transform method of analysis of discrete-time systems parallels the Laplace transform method of analysis of continuous-time systems, with some minor differences. In fact, we shall see that *the z-transform is the Laplace transform in disguise*. The behavior of discrete-time systems is similar to that of continuous-time systems (with some differences). The frequency-domain analysis of discrete-time systems is based on the fact (proved in Sec. 3.8-2) that the response of a linear, time-invariant, discrete-time (LTID) system to an everlasting exponential *zⁿ* is the same exponential (within a multiplicative constant) given by *H*[*z*]*zⁿ*. We then express an input *x*[*n*] as a sum of (everlasting) exponentials of the form *zⁿ*. The system response to *x*[*n*] is then found as a sum of the system's responses to all these exponential components. The tool that allows us to represent an arbitrary input *x*[*n*] as a sum of (everlasting) exponentials of the form *zⁿ* is the *z*-transform. ## **5.1 THE** *z***[-TRANSFORM](#page-11-0)** We define *X*[*z*], the direct *z*-transform of *x*[*n*], as

X[z] = \sum_{n = -\infty}^{\infty} x[n]z^{-n}

$$(5.1) where *z* is a complex variable. The signal *x*[*n*], which is the inverse *z*-transform of *X*[*z*], can be obtained from *X*[*z*] by using the following inverse *z*-transformation:$$

x[n] = \frac{1}{2\pi j} \oint X[z] z^{n-1} dz

$$\n(5.2) The symbol 6 indicates an integration in counterclockwise direction around a closed path in the complex plane (see Fig. 5.1). We derive this *z*-transform pair later, in Ch. 9, as an extension of the discrete-time Fourier transform pair. <span id="page-507-0"></span>CHAPTER **5** As in the case of the Laplace transform, we need not worry about this integral at this point because inverse *z*-transforms of many signals of engineering interest can be found in a *z*-transform table. The direct and inverse *z*-transforms can be expressed symbolically as$$

X[z] = \mathcal{Z}{x[n]} \qquad \text{and} \qquad x[n] = \mathcal{Z}^{-1}{X[z]}

or simply as *x*[*n*] ⇐⇒ *X*[*z*] Note that *^Z*⁻¹ [*Z*{*x*[*n*]}] = *^x*[*n*] and *^Z*[*Z*⁻¹ {*X*[*z*]}] = *X*[*z*] ### LINEARITY OF THE *z*-TRANSFORM Like the Laplace transform, the *z*-transform is a linear operator. If *x*1[*n*] ⇐⇒ *X*1[*z*] and *x*2[*n*] ⇐⇒ *X*2[*z*] then *a*1*x*1[*n*] +*a*2*x*2[*n*] ⇐⇒ *a*1*X*1[*z*] +*a*2*X*2[*z*] The proof is trivial and follows from the definition of the *z*-transform. This result can be extended to finite sums. ### THE UNILATERAL *z*-TRANSFORM For the same reasons discussed in Ch. 4, we find it convenient to consider the unilateral *z*-transform. As seen for the Laplace case, the bilateral transform has some complications because of the non-uniqueness of the inverse transform. In contrast, the unilateral transform has a unique inverse. This fact simplifies the analysis problem considerably, but at a price: the unilateral version can handle only causal signals and systems. Fortunately, most of the practical cases are causal. The more general *bilateral z-transform* is discussed later, in Sec. 5.8. In practice, the term *z-transform* generally means *the unilateral z-transform*. In a basic sense, there is no difference between the unilateral and the bilateral *z*-transform. The unilateral transform is the bilateral transform that deals with a subclass of signals starting at *n* = 0 (causal signals). Hence, the definition of the unilateral transform is the same as that of the bilateral [Eq. (5.1)], except that the limits of the sum are from 0 to ∞:

X[z] = \sum_{n=0}^{\infty} x[n]z^{-n}

(5.3) The expression for the inverse *z*-transform in Eq. (5.2) remains valid for the unilateral case also. ## THE REGION OF CONVERGENCE (ROC) OF *X*[*z*] The sum in Eq. (5.1) [or Eq. (5.3)] defining the direct *z*-transform *X*[*z*] may not converge (exist) for all values of *z*. The values of *z* (the region in the complex plane) for which the sum in Eq. (5.1) converges (or exists) are called the *region of existence,* or more commonly the *region of convergence* (ROC), for *X*[*z*]. This concept will become clear in the following example. ### **EXAMPLE 5.1 Bilateral** *z***-Transform of a Causal Exponential** Find the *z*-transform and the corresponding ROC for the signal γ *nu*[*n*]. By definition,

X[z] = \sum_{n=0}^{\infty} \gamma^n u[n] z^{-n}

$$Since *u*[*n*] = 1 for all *n* ≥ 0,$$

X[z] = \sum_{n=0}^{\infty} \left(\frac{\gamma}{z}\right)^n = 1 + \left(\frac{\gamma}{z}\right) + \left(\frac{\gamma}{z}\right)^2 + \left(\frac{\gamma}{z}\right)^3 + \dots + \dots

$$(5.4) It is helpful to remember the geometric progression and its sum [see Sec. B.8-3]:$$

1 + x + x2 + x3 + \dots = \frac{1}{1 - x} \quad \text{if} \quad |x| < 1

$$Applying this relationship to Eq. (5.4) yields$$

X[z] = \frac{1}{1 - \frac{\gamma}{z}} \qquad \left| \frac{\gamma}{z} \right| < 1

$$\n$$

= \frac{z}{z - \gamma} \qquad |z| > |\gamma| \tag{5.5}

Observe that *X*[*z*] exists only for |*z*| > |γ |. For |*z*| < |γ |, the sum in Eq. (5.4) does not converge; it goes to infinity. Therefore, the ROC of *X*[*z*] is the shaded region outside the circle of radius |γ |, centered at the origin, in the *z*-plane, as depicted in Fig. 5.1b. **Figure 5.1** ^γ *nu*[*n*] and the region of convergence of its *^z*-transform. Later in Eq. (5.52), we show that the *z*-transform of another signal, −γ *nu*[−(*n* + 1)], is also *z*/(*z* − γ ). However, the ROC in this case is |*z*| < |γ |. Clearly, the inverse *z*-transform of *z*/(*z* − γ ) is not unique. However, if we restrict the inverse transform to be causal, then the inverse transform is unique, namely, γ *nu*[*n*]. The ROC is required for evaluating *x*[*n*] from *X*[*z*], according to Eq. (5.2). The integral in Eq. (5.2) is a contour integral, implying integration in a counterclockwise direction along a closed path centered at the origin and satisfying the condition |*z*| > |γ |. Thus, any circular path centered at the origin and with a radius greater than |γ | (Fig. 5.1b) will suffice. We can show that the integral in Eq. (5.2) along any such path (with a radius greater than |γ |) yields the same result, namely, *x*[*n*]. † Such integration in the complex plane requires a background in the theory of functions of complex variables. We can avoid this integration by compiling a table of *z*-transforms (Table 5.1), where *z*-transform pairs are tabulated for a variety of signals. To find the inverse *z*-transform of say, *z*/(*z* − γ ), instead of using the complex integration in Eq. (5.2), we consult the table and find the inverse *z*-transform of *z*/(*z*−γ ) as γ *nu*[*n*]. Because of the uniqueness property of the unilateral *z*-transform, there is only one inverse for each *X*[*z*]. Although the table given here is rather short, it comprises the functions of most practical interest. The situation of the *z*-transform regarding the uniqueness of the inverse transform is parallel to that of the Laplace transform. For the bilateral case, the inverse *z*-transform is not unique unless the ROC is specified. For the unilateral case, the inverse transform is unique; the region of convergence need not be specified to determine the inverse *z*-transform. For this reason, we shall ignore the ROC in the unilateral *z*-transform Table 5.1. ## EXISTENCE OF THE *z*-TRANSFORM By definition,

X[z] = \sum_{n=0}^{\infty} x[n]z^{-n} = \sum_{n=0}^{\infty} \frac{x[n]}{z^n}

$$The existence of the *z*-transform is guaranteed if$$

|X[z]| \leq \sum_{n=0}^{\infty} \frac{|x[n]|}{|z|^n} < \infty

$$for some ^|*z*|. Any signal *^x*[*n*] that grows no faster than an exponential signal *^rⁿ* ⁰, for some *r*0, satisfies this condition. Thus, if$$

|x[n]| \le r_0^n \qquad \text{for some } r_0 \tag{5.6}

$$^† Indeed, the path need not even be circular. It can have any odd shape, as long as it encloses the pole(s) of *X*[*z*] and the path of integration is counterclockwise. | No. | x[n] | X[z] | |-----|----------------------------------------------------|-----------------------------------------------------------------| | 1 | δ[n−k] | z−k | | 2 | u[n] | z<br>z−1 | | 3 | nu[n] | z<br>(z−1)2 | | 4 | n2u[n] | z(z+1)<br>(z−1)3 | | 5 | n3u[n] | z(z2 +4z+1)<br>(z−1)4 | | 6 | γ nu[n] | z<br>z−γ | | 7 | γ n−1u[n−1] | 1<br>z−γ | | 8 | nγ nu[n] | γ z<br>(z−γ )2 | | 9 | n2γ<br>nu[n] | γ z(z+γ )<br>(z−γ )3 | | 10 | n(n−1)(n−2)···(n−m+1)<br>γ nu[n]<br>γ mm! | z<br>(z−γ )m+1 | | 11a | n cos<br> γ <br>βn u[n] | z(z− γ cos β)<br>z2 −(2 γ<br>2<br> cos β)z+ γ | | 11b | n sin<br> γ <br>βn u[n] | z γ sin β<br>z2 −(2 γ<br>2<br> cos β)z+ γ | | 12a | n cos(βn+θ<br>r γ <br>)u[n] | rz[z cos θ − γ cos(β −θ )]<br>z2 −(2 γ<br>2<br> cos β)z+ γ | | 12b | n cos(βn+θ<br>γ = γ ejβ<br>r γ <br>)u[n] | (0.5rejθ )z<br>(0.5re−jθ )z<br>+<br>z−γ<br>z−γ ∗ | | 12c | n cos(βn+θ<br>r γ <br>)u[n] | z(Az+B)<br>z2 +2az+ γ<br>2<br> | | | <br>A2 γ<br>2 +B2 −2AaB<br> <br>r =<br>2 −a2<br> γ | | | | β = cos−1 −a<br> γ | | | | Aa−B<br>θ = tan−1<br><br>2 −a2<br>A<br> γ | | **TABLE 5.1** Select (Unilateral) *z*-Transform Pairs$$

|X[z]| \le \sum_{n=0}^{\infty} \left(\frac{r_0}{|z|}\right)^n = \frac{1}{1 - \frac{r_0}{|z|}} \qquad |z| > r_0

$$Therefore, *X*[*z*] exists for |*z*| > *r*0. Almost all practical signals satisfy Eq. (5.6) and are therefore *z*-transformable. Some signal models (e.g., γ *ⁿ*² ) grow faster than the exponential signal *rⁿ* ⁰ (for any *r*0) and do not satisfy Eq. (5.6) and therefore are not *z*-transformable. Fortunately, such signals are of little practical or theoretical interest. Even such signals over a finite interval are *z*-transformable. then This geometric sum simplifies [see Sec. B.8-3] to$$

X[z] = \frac{1}{1 - \frac{1}{z}} \qquad \left| \frac{1}{z} \right| < 1

$$\n$$

= \frac{z}{z - 1} \qquad |z| > 1

$$Therefore,$$

u[n] \Longleftrightarrow \frac{z}{z-1} \qquad |z| > 1

$$**(c)** Recall that cos β*n* = (*ej*β*ⁿ* +*e*−*j*β*ⁿ*)/2. Moreover, according to Eq. (5.5),$$

e^{\pm j\beta n}u[n] \Longleftrightarrow \frac{z}{z - e^{\pm j\beta}} \qquad |z| > |e^{\pm j\beta}| = 1

$$Therefore,$$

X[z] = \frac{1}{2} \left[ \frac{z}{z - e^{j\beta}} + \frac{z}{z - e^{-j\beta}} \right] = \frac{z(z - \cos \beta)}{z^2 - 2z \cos \beta + 1} \qquad |z| > 1

$$**(d)** Here *x*[0] = *x*[1] = *x*[2] = *x*[3] = *x*[4] = 1 and *x*[5] = *x*[6]=···= 0. Therefore, according to Eq. (5.7),$$

X[z] = 1 + \frac{1}{z} + \frac{1}{z^2} + \frac{1}{z^3} + \frac{1}{z^4} = \frac{z^4 + z^3 + z^2 + z + 1}{z^4}

for all $z \neq 0$ We can also express this result in a more compact form by summing the geometric progression on the right-hand side of the foregoing equation. From the result in Sec. B.8-3 with *r* =1/*z*,*m*= 0, and *n* = 4, we obtain

X[z] = \frac{\left(\frac{1}{z}\right)^5 - \left(\frac{1}{z}\right)^0}{\frac{1}{z} - 1} = \frac{z}{z - 1} (1 - z^{-5})

### **DR ILL 5.1 Bilateral** *z***-Transform** - **(a)** Find the *z*-transform of a signal shown in Fig. 5.3. - **(b)** Use pair 12a (Table 5.1) to find the *z*-transform of *x*[*n*] = 20.65( ^√2)*ⁿ* cos[(π/4)*ⁿ* ⁻ 1.415]*u*[*n*]. <span id="page-514-0"></span> ### **[5.1-1 Inverse Transform by Partial Fraction Expansion and Tables](#page-11-0)** As in the Laplace transform, we shall avoid the integration in the complex plane required to find the inverse *z*-transform [Eq. (5.2)] by using the (unilateral) transform table (Table 5.1). Many of the transforms *X*[*z*] of practical interest are rational functions (ratio of polynomials in *z*), which can be expressed as a sum of partial fractions, whose inverse transforms can be readily found in a table of transform. The partial fraction method works because for every transformable *x*[*n*] defined for *n* ≥ 0, there is a corresponding unique *X*[*z*] defined for |*z*| > *r*⁰ (where *r*⁰ is some constant), and vice versa. ### **EXAMPLE 5.3 Inverse** *z***-Transform by Partial Fraction Expansion** Find the inverse *z*-transforms of (a)

\frac{8z-19}{(z-2)(z-3)}

$$\n(b)$$

\frac{z(2z^2-11z+12)}{(z-1)(z-2)^3}

$$\n(c)$$

\frac{2z(3z+17)}{(z-1)(z^2-6z+25)}

$$**(a)** Expanding *X*[*z*] into partial fractions yields$$

X[z] = \frac{8z - 19}{(z - 2)(z - 3)} = \frac{3}{z - 2} + \frac{5}{z - 3}

$$$$