LTI system difference equation with input

Question

Let's say I am given a difference equation describing an LTI system:

$$y[n] = F(y[n-k],x[n]),$$

where $$x[n] = 2\cos\left(\frac{n\pi}2 + \frac\pi3\right). $$

I know how to find its frequency response and magnitude response. However, I cannot find in my textbook how to find its response to an input x[n] = f(n)... I mean, sure, I can start plugging in numbers and hope for pattern, but I feel like there must be a more clever method. Is there?

Answer 1

With an input $x[n]=z^n$, the output of a system described by ordinary linear difference equations (such as the one you provided) is a scaled version of the input, where the scaling is $H(z)$, in which $H(z)$ is the transfer function. In other words:

$$\bbox[0.5em,#efe,border:0.1em groove navy]{\ x[n]=z^n \Rightarrow y[n]=H(z) z^n\ }\tag{1}$$

In your question the input is $2\cos\left(\frac{n\pi}2 + \frac\pi3\right)$. We need to first convert it to the form $z^n$: $$\begin{align} x[n]&=2\frac{e^{j(\frac{n\pi}2 + \frac\pi3)}+e^{-j(\frac{n\pi}2 + \frac\pi3)}}{2}\\ &=e^{ j\frac\pi3}\left(e^{ j\frac\pi2}\right)^n+e^{ -j\frac\pi3}\left(e^{- j\frac\pi2}\right)^n\\ &=a_1z_1^{n}+a_2z_2^n \end{align}$$ where $z_1=e^{j\frac\pi2}$, $z_2=e^{-j\frac\pi2}$, $a_1=e^{ j\frac\pi3}$ and $a_2=e^{- j\frac\pi3}$.

Using $(1)$, and since the system is linear, the output is: $$\begin{align} y[n]&=a_1H(z_1)z_1^n+a_2H(z_2)z_2^n\\ &=e^{ j\frac\pi3}H(e^{j\frac\pi2})(e^{j\frac\pi2n})+e^{-j\frac\pi3}H(e^{-j\frac\pi2})(e^{-j\frac\pi2n})\tag{2} \end{align}$$

Now calculate $H(z)$ from the difference equation: $$Y(z)=bX(z)-0.81z^{-2}Y(z)\Rightarrow Y(z)(1+0.81z^{-2})=bX(z)$$ $$H(z)=\frac{Y(z)}{X(z)}=\frac{b}{1+0.81z^{-2}}$$ then insert separately $z=z_1=e^{j\frac\pi2}$, and $z=z_2=e^{-j\frac\pi2}$ to find $y[n]$ as in $(2)$

Answer 2

It's important to know that $x[n]=e^{jn\omega_0}$ is an eigenfunction of a discrete-time LTI system:

$$y[n]=\sum_{k=-\infty}^{\infty}h[k]x[n-k]=e^{jn\omega_0}\sum_{k=-\infty}^{\infty}h[k]e^{-jk\omega_0}=e^{jn\omega_0}H(e^{j\omega_0})\tag{1}$$

It's straightforward to derive from $(1)$ the response to an input signal $x[n]=\cos(n\omega_0+\phi)$:

$$y[n]=|H(e^{j\omega_0})|\cos\left(n\omega_0+\phi+\arg\{H(e^{j\omega_0})\}\right)\tag{2}$$

So you only need to compute the expression for the frequency response $H(e^{j\omega})$, and you can use $(2)$ to compute the output corresponding to a sinusoidal input signal.