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A filter has the transfer function
$H(z) = 1 + 2z^{-1} + 4z^{-2} + 8z^{-3}$.
Determine the impulse response of the filter with the modified frequency response
$F(\omega) = H(\omega - 3\pi/8)$.

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1. Original filter: transfer function → impulse response

For a causal LTI discrete–time system, the transfer function in $z$-domain is

$$H(z) = \sum_{n=-\infty}^{\infty} h[n]\,z^{-n}$$

Comparing this with the given

$$H(z) = 1 + 2z^{-1} + 4z^{-2} + 8z^{-3}$$

we can read off the impulse response coefficients directly:

• The coefficient of $z^{0}$ is $1$ → $h[0] = 1$
• The coefficient of $z^{-1}$ is $2$ → $h[1] = 2$
• The coefficient of $z^{-2}$ is $4$ → $h[2] = 4$
• The coefficient of $z^{-3}$ is $8$ → $h[3] = 8$
• No other powers appear → $h[n] = 0$ for $n < 0$ and $n > 3$

So the original impulse response is

$$h[n] = \begin{cases} 1, & n = 0 \\ 2, & n = 1 \\ 4, & n = 2 \\ 8, & n = 3 \\ 0, & \text{otherwise} \end{cases}$$

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2. Frequency response and DTFT reminder

The frequency response of the filter is

$$H(e^{j\omega}) = \sum_{n=-\infty}^{\infty} h[n]\, e^{-j\omega n}$$

This is just the DTFT (discrete-time Fourier transform) of $h[n]$.

Now we’re told that the new filter has a frequency response

$$F(\omega) = H(\omega - 3\pi/8)$$

In standard notation, that means

$$F(e^{j\omega}) = H\big(e^{j(\omega - 3\pi/8)}\big)$$

So it is the original frequency response shifted along the $\omega$-axis by $+3\pi/8$.

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3. Time–frequency shift property

We use the frequency-shift property of the DTFT:

• If $h[n] \xleftrightarrow{\mathcal{F}} H(e^{j\omega})$,
• then the sequence $h[n]\; e^{j\omega_0 n}$ has DTFT $H(e^{j(\omega - \omega_0)})$.

In symbols:

$$h[n] \;\xleftrightarrow{\text{DTFT}}\; H(e^{j\omega})$$ $$h[n]\, e^{j\omega_0 n} \;\xleftrightarrow{\text{DTFT}}\; H(e^{j(\omega - \omega_0)})$$

Compare this to our definition:

$$F(e^{j\omega}) = H(e^{j(\omega - 3\pi/8)})$$

So here, $\omega_0 = 3\pi/8$. That means the new impulse response $f[n]$ is related to the old one $h[n]$ as:

$$f[n] = h[n] \, e^{j(3\pi/8)n}$$

This is the key step: shifting the frequency response by $+3\pi/8$ corresponds to multiplying the time-domain impulse response by $e^{j(3\pi/8)n}$.

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4. Explicit form of the new impulse response

We already know $h[n]$ is non-zero only for $n = 0,1,2,3$. So compute $f[n]$ on these indices:

$$f[n] = h[n]\; e^{j(3\pi/8)n}$$

• For $n = 0$:

  $$f[0] = h[0]\, e^{j(3\pi/8)\cdot 0} = 1 \cdot e^{0} = 1$$

• For $n = 1$:

  $$f[1] = h[1]\, e^{j(3\pi/8)\cdot 1} = 2\, e^{j3\pi/8}$$

• For $n = 2$:

  $$f[2] = h[2]\, e^{j(3\pi/8)\cdot 2} = 4\, e^{j3\pi/4}$$

• For $n = 3$:

  $$f[3] = h[3]\, e^{j(3\pi/8)\cdot 3} = 8\, e^{j9\pi/8}$$

For all other $n$:

$$f[n] = 0,\quad n < 0 \text{ or } n > 3$$

So the impulse response of the modified filter is

$$f[n] = \begin{cases} 1, & n = 0 \\ 2\,e^{j3\pi/8}, & n = 1 \\ 4\,e^{j3\pi/4}, & n = 2 \\ 8\,e^{j9\pi/8}, & n = 3 \\ 0, & \text{otherwise} \end{cases}$$

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5. (Optional) Real and imaginary parts

If you want to see this in terms of cosine and sine, you can write

$$f[n] = h[n]\big(\cos(3\pi n/8) + j\sin(3\pi n/8)\big)$$

Explicitly:

• $f[0] = 1$
• $f[1] = 2\big(\cos(3\pi/8) + j\sin(3\pi/8)\big)$
• $f[2] = 4\big(\cos(3\pi/4) + j\sin(3\pi/4)\big)$
• $f[3] = 8\big(\cos(9\pi/8) + j\sin(9\pi/8)\big)$

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Final answer (compact)

The filter with frequency response $F(\omega) = H(\omega - 3\pi/8)$ has impulse response

$$f[n] = h[n]\,e^{j(3\pi/8)n}$$

where $h[n]$ is the original impulse response,

$$h[n] = \begin{cases} 1, & n=0 \\ 2, & n=1 \\ 4, & n=2 \\ 8, & n=3 \\ 0, & \text{otherwise} \end{cases}$$

So

$$f[n] = \begin{cases} 1, & n = 0 \\ 2\,e^{j3\pi/8}, & n = 1 \\ 4\,e^{j3\pi/4}, & n = 2 \\ 8\,e^{j9\pi/8}, & n = 3 \\ 0, & \text{otherwise} \end{cases}$$