To determine the impulse response $h[n]$ from the given transfer function $H(z)$, we need to find the inverse Z-transform of $H(z)$. Here's how we approach it:

Step 1: Rewrite $H(z)$

The given $H(z)$ is:

$H(z) = \frac{2 + 3z^{-1}}{\left(1 + \frac{1}{2}z^{-1}\right)\left(1 + \frac{1}{4}z^{-1}\right)}.$

Rewriting the denominator in a simpler form:

$H(z) = \frac{2 + 3z^{-1}}{(1 + 0.5z^{-1})(1 + 0.25z^{-1})}.$

Step 2: Expand $H(z)$ into Partial Fractions

We express $H(z)$ as:

$H(z) = \frac{A}{1 + 0.5z^{-1}} + \frac{B}{1 + 0.25z^{-1}},$

where $A$ and $B$ are constants to be determined.

Combine Terms

$H(z) = \frac{A(1 + 0.25z^{-1}) + B(1 + 0.5z^{-1})}{(1 + 0.5z^{-1})(1 + 0.25z^{-1})}.$

Equating the numerator:

$2 + 3z^{-1} = A(1 + 0.25z^{-1}) + B(1 + 0.5z^{-1}).$

Expand and collect terms:

$2 + 3z^{-1} = A + 0.25Az^{-1} + B + 0.5Bz^{-1}.$

Separate constant and $z^{-1}$ terms:

$2 = A + B, \quad 3 = 0.25A + 0.5B.$

Solve for $A$ and $B$

From $2 = A + B$, we have $B = 2 - A$.

Substitute $B = 2 - A$ into the second equation:

$3 = 0.25A + 0.5(2 - A).$

Simplify:

$3 = 0.25A + 1 - 0.5A,$ $3 = 1 - 0.25A,$ $0.25A = -2,$ $A = -8.$

Using $B = 2 - A$:

$B = 2 - (-8) = 10.$

Step 3: Write $H(z)$ with Partial Fractions

Now we rewrite $H(z)$ as:

$H(z) = \frac{-8}{1 + 0.5z^{-1}} + \frac{10}{1 + 0.25z^{-1}}.$

Step 4: Find the Inverse Z-Transform

The general form of the inverse Z-transform for $\frac{1}{1 + az^{-1}}$ is:

[ \mathcal{Z}^{-1}\left[\frac{1}{1 + az^{-1}}\right] = (-a)^n u[n], ]

where $u[n]$ is the unit step function.

Inverse Transform Each Term

For the first term:

[ \mathcal{Z}^{-1}\left[\frac{-8}{1 + 0.5z^{-1}}\right] = -8(-0.5)^n u[n]. ]

For the second term:

[ \mathcal{Z}^{-1}\left[\frac{10}{1 + 0.25z^{-1}}\right] = 10(-0.25)^n u[n]. ]

Step 5: Combine Results

The impulse response $h[n]$ is:

[ h[n] = -8(-0.5)^n u[n] + 10(-0.25)^n u[n}. ]

Or:

$h[n] = (-8)(-0.5)^n + 10(-0.25)^n, \quad n \geq 0.$

Final Answer:

$h[n] = 8(0.5)^n + 10(-0.25)^n, \quad n \geq 0.$

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Do you have questions or want further details? Here are some related questions:

1. How is partial fraction decomposition applied in DSP?
2. What are the properties of the Z-transform that make this calculation possible?
3. How do you interpret the stability of $H(z)$ based on $h[n]$?
4. Can you derive the same $h[n]$ using other methods?
5. What happens if you use $H(z)$ for a different input (e.g., a step input)?

Tip: For verifying impulse responses, you can use numerical tools or directly substitute into the Z-transform formula!