HW 4

• 3-2 Derive a mathematical expression for $H(f)$$H(f)$ and plot $|H(f)|$$|H(f)|$ and $\mathrm{\angle }H(f)$$\angle H(f)$ in Matlab. Draw vertical lines across these plots at each of the frequencies $f_0={\omega }_0/2\pi$$f_0 = \omega_0/2\pi$ given in this problem. Turn in the plots along with your answers.

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• 3-21 Equation (3.30) is labeled (3.29) on page 141.

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• 3-33 The difference equation should read y[n] - (3/4) y[n-1] + (1/8) y[n-2] = ...

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• 3-64 Hint: Express $H(f)=|H(f)|e^{j\mathrm{\angle }H(f)}$$H(f) = |H(f)| e^{j\angle H(f)}$ and then use the chain rule for the derivative.

Hint: Start with the frequency response expressed in terms of its magnitude and phase.

Now differentiate both sides with respect to $\omega$$\omega$.

Next substitute $D(\omega )=-j\frac{d\mathrm{\angle }H(\omega )}{d\omega )}$$D(\omega) = -j\frac{d\angle H(\omega)}{d\omega)}$.

Multiply and divide the first term on the right-hand side by $|H(\omega )|$$|H(\omega)|$.

Next, divide both sides by $H(\omega )$$H(\omega)$.

Now multiply both sides by $j$$j$.

Finally observe that the first term on the right-hand side is pure imaginary and the second term on the right hand side is purely real. Therefore, taking the real part of both sides leads to

Next take the derivative of the DTFT of $h[n]$$h[n]$.

Substituting these into the formula above for $D(\omega )$$D(\omega)$ gives the following.

In the case of the difference equation, we can write the frequency response as

We see that $\mathrm{\angle }H(\omega )=\mathrm{\angle }B(\omega )-\mathrm{\angle }A(\omega )$$\angle H(\omega) = \angle B(\omega) - \angle A(\omega)$. Therefore we can substitute from above to obtain the final answer for this problem.