The Unit Impulse Response

Posted on 2024-05-02 Edited on 2025-06-17 In Electrical Engineering Views: 32

Sources:

B. P. Lathi & Roger Green. (2021). Chapter 2: Time-Domain Analysis of Continuous-Time Systems. Signal Processing and Linear Systems (2nd ed., pp. 163-167). Oxford University Press.

We now discuss a method of determining $h (t)$ , the unit impulse response of an LTIC system described by the $N$ th-order differential equation: $\begin{aligned} (1) & \frac{d^{N} y (t)}{d t^{N}} + \dots + a_{N - 1} \frac{d y (t)}{d t} + a_{N} y (t) \\ = \\ b_{N - M} \frac{d^{M} x (t)}{d t^{M}} + \dots + b_{N - 1} \frac{d x (t)}{d t} + b_{N} x (t) \end{aligned}$

Recall that noise considerations restrict practical systems to $M \leq N$ . Under this constraint, the most general case is $M = N$ . Therefore, it can be expressed as $(D^{N} + a_{1} D^{N - 1} + \dots + a_{N - 1} D + a_{N}) y (t) = (b_{0} D^{N} + b_{1} D^{N - 1} + \dots + b_{N - 1} D + b_{N}) x (t)$

Before deriving the general expression for the unit impulse response $h (t)$ , it is illuminating to understand qualitatively the nature of $h (t)$ .

The impulse response $h (t)$ is the system response to an impulse input $δ (t)$ applied at $t = 0$ with all the initial conditions zero at $t = 0^{-}$ .

An impulse input $δ (t)$ is like lightning, which strikes instantaneously and then vanishes. But in its wake, in that single moment, objects that have been struck are rearranged. Similarly, an impulse input $δ (t)$ appears momentarily at $t = 0$ , and then it is gone forever. But in that moment it generates energy storages; that is, it creates nonzero initial conditions instantaneously within the system at $t = 0^{+}$ .

Although the impulse input $δ (t)$ vanishes for $t > 0$ so that the system has no input after the impulse has been applied, the system will still have a response generated by these newly created initial conditions. The impulse response $h (t)$ , therefore, must consist of the system's characteristic modes for $t \geq 0^{+}$ . As a result, $h (t) = characteristic mode terms t \geq 0^{+}$

The modes can be computed by the zero-input situation!

This response is valid for $t > 0$ . But what happens at $t = 0$ ? At a single moment $t = 0$ , there can at most be an impulse¹, so the form of the complete response $h (t)$ is $h (t) = A_{0} δ (t) + characteristic mode terms t \geq 0$

Setting $x (t) = δ (t)$ and $y (t) = h (t)$ yields $(D^{N} + a_{1} D^{N - 1} + \dots + a_{N - 1} D + a_{N}) h (t) = (b_{0} D^{N} + b_{1} D^{N - 1} + \dots + b_{N - 1} D + b_{N}) δ (t)$

In this equation, we substitute $h (t)$ from Eq. (2.12) and compare the coefficients of similar impulsive terms on both sides:

The highest order of the derivative of impulse on both sides is $N$ , with its coefficient value as $A_{0}$ on the left-hand side and $b_{0}$ on the right-hand side. The two values must be matched.

GREAT TECHNIQUE!

Therefore, $A_{0} = b_{0}$ and $\begin{matrix} (2) & h (t) = b_{0} δ (t) + characteristic modes \end{matrix}$

Note that, in $(1)$ , if $M < N, b_{0} = 0$ .

Hence, the impulse term $b_{0} δ (t)$ exists only if $M = N$ . The unknown coefficients of the $N$ characteristic modes in $h (t)$ in $(2)$ can be determined by using the technique of impulse matching, as explained in the following example.

Simplified Impulse Matching Method

The alternate technique we present now allows us to reduce the procedure to a simple routine to determine $h (t)$ . To avoid the needless distraction, the proof for this procedure is placed in Sec. 2.8. There, we show that for an LTIC system specified by $(1)$ , the unit impulse response $h (t)$ is given by $h (t) = b_{0} δ (t) + [P (D) y_{n} (t)] u (t) .$ where $y_{n} (t)$ is a linear combination of the characteristic modes (remember that, the characteristic modes are computed in the processing of the zero-input response) of the system subject to the following initial conditions: $y_{n} (0) = {\dot{y}}_{n} (0) = {\ddot{y}}_{n} (0) = \dots = y_{n}^{(N - 2)} (0) = 0 and y_{n}^{(N - 1)} (0) = 1$ where $y_{n}^{(k)} (0)$ is the value of the $k$ th derivative of $y_{n} (t)$ at $t = 0$ . We can express this set of conditions for various values of $N$ (the system order) as follows: $\begin{aligned} N = 1 : y_{n} (0) = 1 \\ N = 2 : y_{n} (0) = 0, {\dot{y}}_{n} (0) = 1 \\ N = 3 : y_{n} (0) = {\dot{y}}_{n} (0) = 0, {\ddot{y}}_{n} (0) = 1 \end{aligned}$ and so on.

As stated earlier, if the order of $P (D)$ is less than the order of $Q (D)$ , that is, if $M < N$ , then $b_{0} = 0$ , and the impulse term $b_{0} δ (t)$ in $h (t)$ is zero.

It might be possible for the derivatives of $δ (t)$ to appear at the origin. However, if $M \leq N$ , it is impossible for $h (t)$ to have any derivatives of $δ (t)$ . This conclusion follows from Eq. (2.11) with $x (t) = δ (t)$ and $y (t) = h (t)$ . The coefficients of the impulse and all its derivatives must be matched on both sides of this equation. If $h (t)$ contains $δ^{(1)} (t)$ , the first derivative of $δ (t)$ , the left-hand side of Eq. (2.11) will contain a term $δ^{(N + 1)} (t)$ . But the highest-order derivative term on the right-hand side is $δ^{(N)} (t)$ . Therefore, the two sides cannot match. Similar arguments can be made against the presence of the impulse's higher-order derivatives in $h (t)$ .↩︎