Solution:

As a simple example of how we can find the frequency response of a linear time-invariant system, consider the ideal delay system defined by,

$$ y[n]=x\left[n-n_d\right]...(1) $$

where $n_d$ is a fixed integer. If we consider $x[n]=e^{j \omega n}$ as input to this system, then, from Eq. (1), we have,

$$ y[n]=e^{j \omega\left(n-n_d\right)}=e^{-j \omega n_d} e^{j \omega n} . $$

Thus, for any given value of $\omega$, we obtain an output that is the input multiplied by a complex constant, the value of which depends on the frequency $\omega$ and the delay $n_d$. The frequency response of the ideal delay is therefore

$$ H\left(e^{j \omega}\right)=e^{-j \omega n_d} ....(2) $$

As an alternative method of obtaining the frequency response, recall that $h[n]=$ $\delta\left[n-n_d\right]$ for the ideal delay system. Using Eq. (2.109), we obtain,

$$ H\left(e^{j \omega}\right)=\sum_{n=-\infty}^{\infty} \delta\left[n-n_d\right] e^{-j \omega n}=e^{-j \omega n_d} $$

From the Euler relation, the real and imaginary parts of the frequency response are,

$$ \begin{aligned}\\ & H_R\left(e^{j \omega}\right)=\cos \left(\omega n_d\right)....(3) \\\\ & H_I\left(e^{j \omega}\right)=-\sin \left(\omega n_d\right)...(4)\\ \end{aligned}\\ $$

The magnitude and phase are,

$$ \begin{gathered}\\ \left|H\left(e^{j \omega}\right)\right|=1....(5) \\\\ H\left(e^{j \omega}\right)=-\omega n_d....(6)\\ \end{gathered}\\ $$

We will show that a broad class of signals can be represented as a linear combination of complex exponentials in the form,

$$ x[n]=\sum_k \alpha_k e^{j \omega_k n} . $$

From the principle of superposition, the corresponding output of a linear time-invariant system is,

$$ y[n]=\sum_k \alpha_k H\left(e^{j \omega_k}\right) e^{j \omega_k n} $$