Let a linear, time-invariant, causal discrete linear system be described by the following difference equation:

$$y[n] = 0.9y[n-1] + 0.1x[n]$$

Assuming that $y[-1]=2$ and $x[n]=20\cos(\Omega n)u[n]$, $\Omega = \omega T_s = 0.2\pi$, find the total response, identifiying the terms related to the zero state and zero input responses.

I've found the transfer function of the system to be: $$H(z) = 0.1\frac{z}{z-0.9}$$

and from it I've determined the magnitude and phase of the steady-state response: $$|H \left(e^{j \Omega}\right)| = \frac{0.1}{\sqrt{1.81-1.8\cos(\Omega)}}$$ $$\angle H \left(e^{j \Omega}\right) = -\arctan\left[\frac{-\sin(\Omega)}{1-0.9\cos(\Omega)} \right]$$

Therefore the steady-state response of the system is: $$y_{ss}[n] = 20 \frac{0.1}{\sqrt{1.81-1.8\cos(\Omega)}} \cos\left(\Omega n -\arctan\left[\frac{-\sin(\Omega)}{1-0.9\cos(\Omega)} \right] \right)$$

$$\therefore \boxed{y_{ss}[n] = 3.363\cos(0.2\pi n + 1.138)}$$

But how do I find the total response (zero state and zero input) from the steady-state response?

  • $\begingroup$ why are you assuming $y[-1]=2$? $\endgroup$ Jul 4, 2019 at 5:38
  • $\begingroup$ I don't think he is assuming it, that's part of the problem description. $\endgroup$
    – Max
    Jul 4, 2019 at 5:40
  • $\begingroup$ Max is correct. The condition is given by the problem description. $\endgroup$
    – MCarsten
    Jul 4, 2019 at 5:53

1 Answer 1


Strictly speaking, an LTI system (characterized by an LCCDE) can have a zero-state response, but not a zero-input response. The latter requires nonzero initial conditions which conflicts with the requirement that an LTI system's LCCDE should have zero initial conditions, a.k.a. initial-rest.

However, this problem can be understood in the sense that the given LCCDE represents an LTI system under initial-rest assumptions, yet we still want to compute its response due to nonzero initial conditions as well, in which case the LCCDE actually represents a non LTI system.

Then you can proceed as follows: Lets use one-sided (unilateral) Z-transform to account for the initial conditions: $$ X(z) = \sum_{n=0}^{\infty} x[n] z^{-n} \tag{1} $$

and its useful property: $$ x[n-1] \longleftrightarrow z^{-1}X(z) + x[-1] \tag{2}$$

and for general shift of k: $$ x[n-k] \longleftrightarrow z^{-k} X(z) + \sum_{m=0}^{k-1} z^{-m}x[m-k] \tag{3} $$

Given an N-th order LCCDE: $$ \sum_{k=0}^{N} a_k y[n-k] = \sum_{k=0}^{M} b_k x[n-k] \tag{4} $$

Apply Eq.3 to Eq.4, assuming $x[n]=0, n< 0$:

$$ \sum_{k=0}^{N} a_k \left( z^{-k} Y(z) + \sum_{m=0}^{k-1} z^{-m} y[m-k] \right) = \sum_{k=0}^{M} b_k z^{-k} X(z) \tag{5} $$

$$ Y(z) \left( \sum_{k=0}^{N} a_k z^{-k} \right) + \sum_{k=0}^{N} a_k \sum_{m=0}^{k-1} z^{-m} y[m-k] = X(z) \left(\sum_{k=0}^{M} b_k z^{-k} \right) \tag{6} $$

Denote the double sum $\sum_{k=0}^{N} a_k \sum_{m=0}^{k-1} z^{-m} y[m-k]$ as $Y_{ic}(z)$ the initial-conditions part of output, then Eq.6 becomes: $$ Y(z) \left( \sum_{k=0}^{N} a_k z^{-k} \right) + Y_{ic}(z) = X(z) \left(\sum_{k=0}^{M} b_k z^{-k} \right) \tag{7} $$

Then express $Y(z)$ as: $$ Y(z) = \frac{ \sum_{k=0}^{M} b_k z^{-k}}{\sum_{k=0}^{N} a_k z^{-k}} X(z) - \frac{ Y_{ic}(z) }{\sum_{k=0}^{N} a_k z^{-k} } \tag{8} $$

Denoting $\frac{ \sum_{k=0}^{M} b_k z^{-k}}{\sum_{k=0}^{N} a_k z^{-k}}$ as $H(z)$; the Transfer Function of the associated LTI system, we have:

$$ Y(z) = H(z) X(z) - \frac{ Y_{ic}(z) }{\sum_{k=0}^{N} a_k z^{-k} } \tag{9} $$

Eq.9 expresses the LCCDE output $Y(z)$ in terms of the input $X(z)$ and the initial-conditions $Y_{ic}(z)$.

The first term $H(z) X(z)$ is the zero-state response (or forced response) $Y_{zs}(z)$ corresponding to the input $X(z)$, and the second term is the zero-input response (or natural response) $Y_{zi}(z)$ corresponding to the initial-conditions $Y_{ic}(z)$ which is typically a transient response due to initial conditions.

Hence we have:

$$Y(z) = Y_{zs}(z) + Y_{zi}(z) \tag{10}$$

The zero-state response, $Y_{zs}(z) = H(z)X(z)$ corresponds to the LTI system that's characterized by the LCCDE, because it assumes zero initial conditions;i.e., $Y_{IC}(z) = 0$, and then the LCCDE output is given by :

$$ Y(z) = H(z)X(s) \longleftrightarrow y[n] = h[n] \star x[n] \tag{11} $$

Eq.11 displays the equivalence between the zero-state response of an LCCDE and the convolution output of the corresponding LTI system with impulse response $h[n]$.

For a suddenly applied sinudoidal $x[n] = K \cos(\omega_0 n)u[n]$, the zero-state response will include two terms: a switching transient part that corresponds to the sudden application of the input; i.e. $x[n] = 0 , n < 0$ , and a sinusoidal steady-state part that remains persistent after switching transients (and initial-condition transients as well, if that also exists) decay, and which corresponds to the response of the LTI system as if the input were $x[n] = \cos(\omega_0 n)$.

When applied to your LCCDE, you get the following:

$$ y[n] + a y[n-1] = b_0 x[n] ~~~,~~~ y[-1]=2 \tag{12} $$


$$x[n] = K \cos(\omega_0 n)u[n] \leftrightarrow \frac{K/2}{ 1 - e^{j \omega_0} z^{-1} } + \frac{K/2}{ 1 - e^{-j \omega_0} z^{-1} } \tag{13}$$

where $a=-0.9 ,~ b_0 = 0.1 ,~ \omega_0 = 0.2 \pi$ , and $K=20$.

Then the total output becomes:

$$ Y(z) = \frac{b_0}{1 + a z^{-1}} \left( \frac{K / 2} {1 - e^{j \omega_0} z^{-1}} + \frac{K / 2} {1 - e^{-j \omega_0} z^{-1}} + \right) - \frac{ a y[-1]}{1 + a z^{-1}} \tag{14}$$

Where the term in the paranthesis corresponds to the zero-state output, and the remaing last term is the zero-input response.

Apply partial fraction expansion, PFE, to the zero-state response part:

$$Y(z) = \frac{A_1}{1 - e^{j\omega_0} z^{-1}} + \frac{A_2}{1 - e^{-j\omega_0} z^{-1}} + \frac{B_1}{1 + a z^{-1}} + \frac{B_2}{1 + a z^{-1}} - \frac{a y[-1]}{1 + a z^{-1}}\tag{15} $$

It will turn out that $A_2 = A_1^*$ and $B_2 = B_1^*$, and we have $$A = \frac{b_0 K/2}{1 + a e^{-j \omega_0}} ~~ ,~~ B = \frac{b_0 K/2}{1 + a^{-1} e^{j\omega_0}} $$.

Finally in the time-domain, the following is obtained for the total output:

$$y[n] = 2|A| \cos(\omega_0 n + \phi_A) + 2 \mathcal{Re}\{B\} (-a)^n - a y[-1]( -a)^n \tag{16} $$

For the given values it happens to be:

$$y[n] = 3.3626 \cos(0.2\pi n - 1.0960) u[n] + 0.4629 (0.9)^n u[n] + 1.8 (0.9)^n u[n] \tag{17} $$

Finally we can denote the parts of total output as follows:

the sinusoidal steady-state response: $$y_{ss}[n] = 3.3626 \cos(0.2\pi n - 1.0960) u[n] $$

the input switching transient response: $$y_{is}[n] = 0.4629 (0.9)^n u[n] $$

the zero-state response: $$y_{zs}[n] = y_{ss}[n] + y_{is}[n] $$

the zero-input (transient) response: $$ y_{zi}[n] = 1.8 (0.9)^n u[n] $$


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