FREQUENCY RESPONSE OF DISCRETE-TIME SYSTEMS

5.5 FREQUENCY RESPONSE OF DISCRETE-TIME SYSTEMS

For (asymptotically or BIBO-stable) continuous-time systems, we showed that the system response to an input e j ωt is H(j ω)e j ωt and For (asymptotically or BIBO-stable) continuous-time systems, we showed that the system response to an input e j ωt is H(j ω)e j ωt and

The proof is similar to the one used for continuous-time systems. In Section 3.8-3 , we showed that an LTID system response to an (everlasting) exponential z n is also an (everlasting) exponential H[z]z n. This result is valid only for values of z for which H[z], as defined in Eq. (5.14a) , exists (converges). As usual, we represent this input-output relationship by a directed arrow notation as

Setting z = e j Ω in this relationship yields

Noting that cos Ωn is the real part of e j Ωn , use of Eq. (3.66b) yields

Expressing H[e j Ω ] in the polar form

Eq. (5.44) can be expressed as In other words, the system response y[n] to a sinusoidal input cos Ωn is given by

Following the same argument, the system response to a sinusoid cos ( Ωn + θ) is This result is valid only for BIBO-stable or asymptotically stable systems. The frequency response is meaningless for BIBO-unstable

systems (which include marginally stable and asymptotically unstable systems). This follows from the fact that the frequency response in Eq. (5.43) is obtained by setting z = e j Ω in Eq. (5.42) . But, as shown in Section 3.8-3 [ Eqs. (3.71) ], the relationship (5.42) applies only for values of z for which H[z] exists. For BIBO unstable systems, the ROC for H[z] does not include the unit circle where z = e j Ω. This means, for BIBO-unstable systems, that H[z] is meaningless when z = e j Ω[ † ]

This important result shows that the response of an asymptotically or BIBO-stable LTID system to a discrete-time sinusoidal input of frequency Ω is also a discrete-time sinusoid of the same frequency. The amplitude of the output sinusoid is |H[e j Ω ]| times the input amplitude, and the phase of the output sinusoid is shifted by ∠H[e j Ω ] with respect to the input phase. Clearly |H[e j Ω ]| is the amplitude gain, and a plot of |H[e j Ω ]| versus Ω is the amplitude response of the discrete-time system. Similarly, ∠H[e j Ω ] is the phase response of the system, and a plot of ∠H[e j Ω ] versus Ω shows how the system modifies or shifts the phase of the input sinusoid. Note that H[e j Ω ] incorporates the information of both amplitude and phase responses and therefore is called the frequency responses of the system.

STEADY-STATE RESPONSE TO CAUSAL SINUSOIDAL INPUT As in the case of continuous-time systems, we can show that the response of an LTID system to a causal sinusoidal input cos Ωn u[n]

is y[n] in Eq. (5.46a) , plus a natural component consisting of the characteristic modes (see Prob. 5.5-6). For a stable system, all the modes decay exponentially, and only the sinusoidal component in Eq. (5.46a) persists. For this reason, this component is called the sinusoidal steady-state response of the system. Thus, y ss [n], the steady-state response of a system to a causal sinusoidal input cos

Ωn u[n], is SYSTEM RESPONSE TO SAMPLED CONTINUOUS-TIME SINUSOIDS

So far we have considered the response of a discrete-time system to a discrete-time sinusoid cos Ωn (or exponential e j Ωn ). In practice, the input may be a sampled continuous-time sinusoid cos ωt (or an exponential e j ωt ). When a sinusoid cos ωt is sampled with sampling interval T, the resulting signal is a discrete-time sinusoid cos ωnT, obtained by setting t = nT in cos ωt. Therefore, all the

results developed in this section apply if we substitute ωT for Ω:

EXAMPLE 5.10

For a system specified by the equation find the system response to the input

a. 1 n =1

b.

c. a sampled sinusoid cos 1500t with sampling interval T = 0.001 The system equation can be expressed as

Therefore, the transfer function of the system is

The frequency response is

Therefore

and

The amplitude response |H[e j Ω ]| can also be obtained by observing that |H| 2 =HH*. Therefore

From Eq. (5.48) it follows that

which yields the result found earlier in Eq. (5.49a) Figure 5.14 shows plots of amplitude and phase response as functions of Ω. We now compute the amplitude and the phase response

for the various inputs.

a. x[n] = 1 n =1 Since 1 n =(e j Ω ) n with Ω = 0, the amplitude response is H[e j0 ]. From Eq. (5.49a) we obtain

Therefore

These values also can be read directly from Fig. 5.14a and 5.14b , respectively, corresponding to Ω = 0. Therefore, the system response to input 1 is

b. x[n] = cos[( π/6)n −0.2] Here Ω = π/6. According to Eqs. (5.49)

These values also can be read directly from Fig. 5.14a and 5.14b , respectively, corresponding to Ω=π/6. Therefore

Figure 5.15 shows the input x[n] and the corresponding system response.

Figure 5.15: Sinusoidal input and the corresponding output of the LTID system.

c. A sinusoid cos 1500t sampled every T seconds (t = nT) results in a discrete-time sinusoid

For T = 0.001, the input is In this case, Ω =1.5. According to Eqs. (5.49a) and (5.49b) ,

These values also could be read directly from Fig. 5.14 corresponding to Ω =1.5. Therefore

Figure 5.14: Frequency response of the LTID system.