(c)

← Back to LINEAR SYSTEMS AND SIGNALS Overview (c)

$$|z| = \sqrt{(-2)^2 + (-3)^2} = \sqrt{13}$$

$\angle z = \tan^{-1}\left(\frac{-3}{-2}\right) = -123.7^{\circ}$

In this case the angle appears in the third quadrant (see Fig. B.4c), and therefore the answer obtained by the calculator (tan−1(−3/−2) = 56.3◦) is off by 180◦. The correct answer is (56.3 ± 180)◦ = 236.3◦ or −123.7◦. We choose the principal value −123.7◦ so that (see Fig. B.4c)

$$-2 - j3 = \sqrt{13}e^{-j123.7^{\circ}}$$

(d)

$$|z| = \sqrt{1^2 + (-3)^2} = \sqrt{10}$$

$\angle z = \tan^{-1}\left(\frac{-3}{1}\right) = -71.6^{\circ}$

In this case the angle appears in the fourth quadrant (see Fig. B.4d), and therefore the answer given by the calculator, tan−¹(−3/1) = −71.6◦, is correct (see Fig. B.4d):

$$1 - j3 = \sqrt{10}e^{-j71.6^{\circ}}$$

10 CHAPTER B BACKGROUND

We can easily verify these results using the MATLAB abs and angle commands. To obtain units of degrees, we must multiply the radian result of the angle command by 180 ^π . Furthermore, the angle command correctly computes angles for all four quadrants of √ the complex plane. To provide an example, let us use MATLAB to verify that −2 + j1 = 5*e^j153.4◦ = 2.2361e^j*153.4◦ .

abs(-2+1j) ans = 2.2361 >> angle(-2+1j)*180/pi ans = 153.4349

One can also use the cart2pol command to convert Cartesian to polar coordinates. Readers, particularly those who are unfamiliar with MATLAB, will benefit by reading the overview in Sec. B.7.

EXAMPLE B.2 Polar to Cartesian Form

Represent the following numbers in the complex plane and express them in Cartesian form: (a) 2ejπ/3, (b) 4e−j3π/4, (c) 2ejπ/2, (d) 3e−j3^π , (e) 2ej4^π , and (f) 2e−j4^π .

(a) 2ejπ/³ = 2(cos π/3+jsin π/3) = 1+j ^√3 (see Fig. B.5a) (b) 4e−j3π/⁴ = 4(cos 3π/4−jsin 3π/4) = −2 ^√2−j^{2 √}2 (see Fig. B.5b) (c) 2ejπ/² = 2(cos π/2+jsin π/2) = 2(0+j1) = j2 (see Fig. B.5c) (d) 3e−j3^π = 3(cos 3π −jsin 3π ) = 3(−1+j0) = −3 (see Fig. B.5d) (e) 2ej4^π = 2(cos 4π +jsin 4π ) = 2(1+j0) = 2 (see Fig. B.5e) (f) 2e−j4^π = 2(cos 4π −jsin 4π ) = 2(1−j0) = 2 (see Fig. B.5f)

We can readily verify these results using MATLAB. First, we use the exp function to represent a number in polar form. Next, we use the real and imag commands to determine the real and imaginary components of that number. To provide an example, let us use MATLAB to verify the result of part (a): 2ejπ/³ = 1+j ^{√3 = 1}+j1.7321.

>> real(2*exp(1j*pi/3))
    ans = 1.0000
>> imag(2*exp(1j*pi/3))
    ans = 1.7321

Since MATLAB defaults to Cartesian form, we could have verified the entire result in one step.

>> 2*exp(1j*pi/3)
    ans = 1.0000 + 1.7321i

One can also use the pol2cart command to convert polar to Cartesian coordinates.

ARITHMETICAL OPERATIONS, POWERS, AND ROOTS OF COMPLEX NUMBERS

To conveniently perform addition and subtraction, complex numbers should be expressed in Cartesian form. Thus, if

$$z_1 = 3 + j4 = 5e^{j53.1^{\circ}}$$

and

$$z_2 = 2 + j3 = \sqrt{13}e^{j56.3^{\circ}}$$

then

$$z_1 + z_2 = (3 + j4) + (2 + j3) = 5 + j7$$

12 CHAPTER B BACKGROUND

If z¹ and z² are given in polar form, we would need to convert them into Cartesian form for the purpose of adding (or subtracting). Multiplication and division, however, can be carried out in either Cartesian or polar form, although the latter proves to be much more convenient. This is because if z¹ and z² are expressed in polar form as

$$z_1 = r_1 e^{j\theta_1}$$

and $z_2 = r_2 e^{j\theta_2}$

then

$$z_1 z_2 = (r_1 e^{j\theta_1})(r_2 e^{j\theta_2}) = r_1 r_2 e^{j(\theta_1 + \theta_2)}$$

and

$$\frac{z_1}{z_2} = \frac{r_1 e^{j\theta_1}}{r_2 e^{j\theta_2}} = \frac{r_1}{r_2} e^{j(\theta_1 - \theta_2)}$$

Moreover,

$$z^n = (re^{j\theta})^n = r^n e^{jn\theta}$$

and

$$z^{1/n} = (re^{i\theta})^{1/n} = r^{1/n}e^{i\theta/n}$$

(B.11)

This shows that the operations of multiplication, division, powers, and roots can be carried out with remarkable ease when the numbers are in polar form.

Strictly speaking, there are n values for z1/ⁿ (the nth root of z). To find all the n roots, we reexamine Eq. (B.11):

$$z^{1/n} = [re^{j\theta}]^{1/n} = [re^{j(\theta + 2\pi k)}]^{1/n} = r^{1/n}e^{j(\theta + 2\pi k)/n} \qquad k = 0, 1, 2, \dots, n-1$$

(B.12)

The value of z1/ⁿ given in Eq. (B.11) is the principal value of z1/ⁿ, obtained by taking the nth root of the principal value of z, which corresponds to the case k = 0 in Eq. (B.12).

EXAMPLE B.3 Multiplication and Division of Complex Numbers

Using both polar and Cartesian forms, determine z1z² and z1/z² for the numbers

$$z_1 = 3 + j4 = 5e^{j53.1^{\circ}}$$

and $z_2 = 2 + j3 = \sqrt{13}e^{j56.3^{\circ}}$

Multiplication: Cartesian Form

$$z_1 z_2 = (3+j4)(2+j3) = (6-12) + j(8+9) = -6+j17$$

Multiplication: Polar Form

$$z_1 z_2 = (5e^{j53.1^{\circ}})(\sqrt{13}e^{j56.3^{\circ}}) = 5\sqrt{13}e^{j109.4^{\circ}}$$

Division: Cartesian Form

$$\frac{z_1}{z_2} = \frac{3+j4}{2+j3}$$

To eliminate the complex number in the denominator, we multiply both the numerator and the denominator of the right-hand side by 2−j3, the denominator’s conjugate. This yields

$$\frac{z_1}{z_2} = \frac{(3+j4)(2-j3)}{(2+j3)(2-j3)} = \frac{18-j1}{2^2+3^2} = \frac{18-j1}{13} = \frac{18}{13} - j\frac{1}{13}$$

Division: Polar Form

$$\frac{z_1}{z_2} = \frac{5e^{j53.1^{\circ}}}{\sqrt{13}e^{j56.3^{\circ}}} = \frac{5}{\sqrt{13}}e^{j(53.1^{\circ} - 56.3^{\circ})} = \frac{5}{\sqrt{13}}e^{-j3.2^{\circ}}$$

It is clear from this example that multiplication and division are easier to accomplish in polar form than in Cartesian form.

These results are also easily verified using MATLAB. To provide one example, let us use Cartesian forms in MATLAB to verify that z1z² = −6+j17.

>> z1 = 3+4j; z2 = 2+3j;
>> z1*z2
    ans = -6.0000 + 17.0000i

As a second example, let us use polar forms in MATLAB to verify that z1/z² = 1.3868e−j3.2◦ . Since MATLAB generally expects angles be represented in the natural units of radians, we must use appropriate conversion factors in moving between degrees and radians (and vice versa).

>> z1 = 5*exp(1j*53.1*pi/180); z2 = sqrt(13)*exp(1j*56.3*pi/180);
>> abs(z1/z2)
    ans = 1.3868
>> angle(z1/z2)*180/pi
    ans = -3.2000

EXAMPLE B.4 Working with Complex Numbers

For z¹ = 2ejπ/⁴ and z² = 8ejπ/3, find the following: (a) 2z¹ −z2, (b) 1/z1, (c) z1/z^{2 2}, and (d) ^√3 *^z*2.

(a) Since subtraction cannot be performed directly in polar form, we convert z¹ and z² to Cartesian form:

$$z_1 = 2e^{j\pi/4} = 2\left(\cos\frac{\pi}{4} + j\sin\frac{\pi}{4}\right) = \sqrt{2} + j\sqrt{2}$$ $$z_2 = 8e^{j\pi/3} = 8\left(\cos\frac{\pi}{3} + j\sin\frac{\pi}{3}\right) = 4 + j4\sqrt{3}$$

Therefore,

$$2z_1 - z_2 = 2(\sqrt{2} + j\sqrt{2}) - (4 + j4\sqrt{3}) = (2\sqrt{2} - 4) + j(2\sqrt{2} - 4\sqrt{3}) = -1.17 - j4.1$$

(b)

$$\frac{1}{z_1} = \frac{1}{2e^{j\pi/4}} = \frac{1}{2}e^{-j\pi/4}$$

(c)

$$\frac{z_1}{z_2^2} = \frac{2e^{j\pi/4}}{(8e^{j\pi/3})^2} = \frac{2e^{j\pi/4}}{64e^{j2\pi/3}} = \frac{1}{32}e^{j(\pi/4 - 2\pi/3)} = \frac{1}{32}e^{-j(5\pi/12)}$$

(d) There are three cube roots of 8ej(π/3) = 8ej(π/3+2πk) , k = 0, 1, 2.

$$\sqrt[3]{z_2} = z_2^{1/3} = \left[8e^{i(\pi/3 + 2\pi k)}\right]^{1/3} = 8^{1/3} \left(e^{i[(6\pi k + \pi)/3]}\right)^{1/3} = \begin{cases} 2e^{i\pi/9} & k = 0\\ 2e^{i7\pi/9} & k = 1\\ 2e^{i13\pi/9} & k = 2 \end{cases}$$

The value corresponding to k = 0 is termed the principal value.

EXAMPLE B.5 Standard Forms of Complex Numbers

Consider X(ω), a complex function of a real variable ω:

$$X(\omega) = \frac{2 + j\omega}{3 + j4\omega}$$

(a) Express X(ω) in Cartesian form, and find its real and imaginary parts.

(b) Express X(ω) in polar form, and find its magnitude |X(ω)| and angle X(ω).

$$X(\omega) = \frac{(2+j\omega)(3-j4\omega)}{(3+j4\omega)(3-j4\omega)} = \frac{(6+4\omega^2) - j5\omega}{9+16\omega^2} = \frac{6+4\omega^2}{9+16\omega^2} - j\frac{5\omega}{9+16\omega^2}$$

This is the Cartesian form of X(ω). Clearly, the real and imaginary parts Xr(ω) and Xi(ω) are given by

$$X_r(\omega) = \frac{6 + 4\omega^2}{9 + 16\omega^2}$$

and $X_i(\omega) = \frac{-5\omega}{9 + 16\omega^2}$

⁽a) To obtain the real and imaginary parts of X(ω), we must eliminate imaginary terms in the denominator of X(ω). This is readily done by multiplying both the numerator and the denominator of X(ω) by 3−j4ω, the conjugate of the denominator 3+j4ω so that

$$(\mathbf{b})$$ $$X(\omega) = \frac{2 + j\omega}{3 + j4\omega} = \frac{\sqrt{4 + \omega^2} e^{j\tan^{-1}(\omega/2)}}{\sqrt{9 + 16\omega^2} e^{j\tan^{-1}(4\omega/3)}} = \sqrt{\frac{4 + \omega^2}{9 + 16\omega^2}} e^{j\tan^{-1}(\omega/2) - \tan^{-1}(4\omega/3)}$$

This is the polar representation of X(ω). Observe that

$$|X(\omega)| = \sqrt{\frac{4 + \omega^2}{9 + 16\omega^2}} \quad \text{and} \quad \angle X(\omega) = \tan^{-1}\left(\frac{\omega}{2}\right) - \tan^{-1}\left(\frac{4\omega}{3}\right)$$

LOGARITHMS OF COMPLEX NUMBERS

To take the natural logarithm of a complex number z, we first express z in general polar form as

$$z = re^{j\theta} = re^{j(\theta \pm 2\pi k)}$$

$k = 0, 1, 2, 3, ...$

Taking the natural logarithm, we see that

$$\ln z = \ln \left( r e^{j(\theta \pm 2\pi k)} \right) = \ln r \pm j(\theta + 2\pi k) \qquad k = 0, 1, 2, 3, \dots$$

The value of lnz for k = 0 is called the principal value of lnz and is denoted by Lnz. In this way, we see that

$$\ln 1 = \ln(1e^{\pm j2\pi k}) = \pm j2\pi k \qquad k = 0, 1, 2, 3, \dots$$ $$\ln(-1) = \ln[1e^{\pm j\pi(2k+1)}] = \pm j(2k+1)\pi \qquad k = 0, 1, 2, 3, \dots$$ $$\ln j = \ln(e^{j\pi(1\pm 4k)/2}) = j\frac{\pi(1\pm 4k)}{2} \qquad k = 0, 1, 2, 3, \dots$$ $$j^j = e^{j\ln j} = e^{-\pi(1\pm 4k)/2} \qquad k = 0, 1, 2, 3, \dots$$

In all of these cases, setting k = 0 yields the principal value of the expression.

We can further our logarithm skills by noting that the familiar properties of logarithms hold for complex arguments. Therefore, we have

$$log(z1z2) = log z1 + log z2$$ $$log(z1/z2) = log z1 - log z2$$ $$a(z1+z2) = az1 × az2$$ $$zc = ecln z$$ $$az = ezln a$$