[2.8 APPENDIX: DETERMINING THE](#page-9-0) IMPULSE RESPONSE

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2.8 APPENDIX: DETERMINING THE IMPULSE RESPONSE

In Eq. (2.13), we showed that for an LTIC system S specified by Eq. (2.11), the unit impulse response h(t) can be expressed as

$$h(t) = b_0 \delta(t) + \text{characteristic modes} \tag{2.51}$$

To determine the characteristic mode terms in Eq. (2.51), let us consider a system S⁰ whose input x(t) and the corresponding output w(t) are related by

$$Q(D)w(t) = x(t) \tag{2.52}$$

Observe that both the systems S and S⁰ have the same characteristic polynomial; namely, Q(λ), and, consequently, the same characteristic modes. Moreover, S⁰ is the same as S with P(D)=1, that is, b⁰ = 0. Therefore, according to Eq. (2.51), the impulse response of S⁰ consists of characteristic mode terms only without an impulse at t = 0. Let us denote this impulse response of S⁰ by yn(t). Observe that yn(t) consists of characteristic modes of S and therefore may be viewed as a zero-input response of S. Now yn(t) is the response of S⁰ to input δ(t). Therefore, according to Eq. (2.52),

$$Q(D)y_n(t) = \delta(t)$$

or

$$(D^N + a_1 D^{N-1} + \dots + a_{N1} D + a_N) y_n(t) = \delta(t)$$

or

$$y_n^{(N)}(t) + a_1 y_n^{(N-1)}(t) + \dots + a_{N-1} y_n^{(1)}(t) + a_N y_n(t) = \delta(t)$$

where y(k) ⁿ (t) represents the kth derivative of yn(t). The right-hand side contains a single impulse term, δ(t). This is possible only if y(N−1) ⁿ (t) has a unit jump discontinuity at t = 0, so that y(N) ⁿ (t) = δ(t). Moreover, the lower-order terms cannot have any jump discontinuity because this would mean the presence of the derivatives of δ(t). Therefore yn(0) = y(1) ⁿ (0) =···= y(N−2) ⁿ (0) = 0 (no discontinuity at t = 0), and the N initial conditions on yn(t) are

$$y_n(0) = y_n^{(1)}(0) = \dots = y_n^{(N-2)}(0) = 0$$

and $y_n^{(N-1)}(0) = 1$ (2.53)

This discussion means that yn(t) is the zero-input response of the system S subject to initial conditions [Eq. (2.53)].

We now show that for the same input x(t) to both systems, S and S0, their respective outputs y(t) and w(t) are related by

$$y(t) = P(D)w(t)$$

\n^(2.54)

To prove this result, we operate on both sides of Eq. (2.52) by P(D) to obtain

$$Q(D)P(D)w(t) = P(D)x(t)$$

Comparison of this equation with Eq. (2.2) leads immediately to Eq. (2.54).

Now if the input x(t) = δ(t), the output of S⁰ is yn(t), and the output of S, according to Eq. (2.54), is P(D)yn(t). This output is h(t), the unit impulse response of S. Note, however, that because it is an impulse response of a causal system S0, the function yn(t) is causal. To incorporate this fact we must represent this function as yn(t)u(t). Now it follows that h(t), the unit impulse response of the system S, is given by

$$h(t) = P(D)[y_n(t)u(t)]$$

\n

$$(2.55)$$

where yn(t) is a linear combination of the characteristic modes of the system subject to initial conditions (2.53).

The right-hand side of Eq. (2.55) is a linear combination of the derivatives of yn(t)u(t). Evaluating these derivatives is clumsy and inconvenient because of the presence of u(t). The derivatives will generate an impulse and its derivatives at the origin. Fortunately when M ≤ N [Eq. (2.11)], we can avoid this difficulty by using the observation in Eq. (2.51), which asserts that at t = 0 (the origin), h(t) = b0δ(t). Therefore, we need not bother to find h(t) at the origin. This simplification means that instead of deriving P(D)[yn(t)u(t)], we can derive P(D)yn(t) and add to it the term b0δ(t) so that

$$h(t) = b_0 \delta(t) + P(D) y_n(t) \qquad t \ge 0$$

= $b_0 \delta(t) + [P(D) y_n(t)] u(t)$

This expression is valid when M ≤ N [the form given in Eq. (2.11)]. When M > N, Eq. (2.55) should be used.