The causal signal with z-transform z 2(z - a) -2 is (u[n] is the unit step signal)
(n + 1) a nu[n]
This problem asks us to identify the causal signal whose Z-transform is given as $X(z) = \frac{z^2}{(z-a)^2}$. We need to work backward from the Z-transform to find the corresponding time-domain signal $x[n]$. The signal is causal, meaning it is zero for $n < 0$, which is usually indicated by the presence of the unit step function, $u[n]$.
Finding the Causal Signal from Z-transform
We can use known Z-transform pairs and properties to solve this. A fundamental Z-transform pair is:
$$ \mathcal{Z}\{a^n u[n]\} = \frac{z}{z-a} $$We also use the multiplication by $n$ property of Z-transforms, which states that if $\mathcal{Z}\{x[n]\} = X(z)$, then:
$$ \mathcal{Z}\{n x[n]\} = -z \frac{dX(z)}{dz} $$Let's start with the signal $a^n u[n]$ and apply the properties to see if we can reach the given transform.
Step 1: Transform of $a^n u[n]$
As mentioned, the Z-transform of $a^n u[n]$ is:
$$ X_1(z) = \mathcal{Z}\{a^n u[n]\} = \frac{z}{z-a} $$Step 2: Applying the multiplication by 'n' property
Now, let's find the Z-transform of the signal $n a^n u[n]$ using the property mentioned above. Here, $x[n] = a^n u[n]$ and $X(z) = \frac{z}{z-a}$.
We need to calculate the derivative of $X(z)$ with respect to $z$:
$$ \frac{dX_1(z)}{dz} = \frac{d}{dz} \left( \frac{z}{z-a} \right) $$Using the quotient rule $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$:
$$ \frac{d}{dz} \left( \frac{z}{z-a} \right) = \frac{(1)(z-a) - (z)(1)}{(z-a)^2} = \frac{z - a - z}{(z-a)^2} = \frac{-a}{(z-a)^2} $$Now, apply the multiplication by $n$ property:
$$ \mathcal{Z}\{n a^n u[n]\} = -z \frac{dX_1(z)}{dz} = -z \left( \frac{-a}{(z-a)^2} \right) = \frac{az}{(z-a)^2} $$Let's call this $X_2(z)$.
Step 3: Relating to the target Z-transform
The target Z-transform is $X(z) = \frac{z^2}{(z-a)^2}$. We found that $\mathcal{Z}\{n a^n u[n]\} = \frac{az}{(z-a)^2}$.
Notice that $X(z) = \frac{z^2}{(z-a)^2} = \frac{1}{a} \left( \frac{az}{(z-a)^2} \right) z$. This relationship isn't directly helpful. Let's rethink.
Consider the signal $(n+1) a^n u[n]$. This can be written as $n a^n u[n] + a^n u[n]$.
The Z-transform of $n a^n u[n]$ is $\frac{az}{(z-a)^2}$.
The Z-transform of $a^n u[n]$ is $\frac{z}{z-a}$.
Using the linearity property of the Z-transform ($\mathcal{Z}\{x_1[n] + x_2[n]\} = X_1(z) + X_2(z)$):
$$ \mathcal{Z}\{(n+1) a^n u[n]\} = \mathcal{Z}\{n a^n u[n]\} + \mathcal{Z}\{a^n u[n]\} $$ $$ = \frac{az}{(z-a)^2} + \frac{z}{z-a} $$To add these fractions, we find a common denominator:
$$ = \frac{az}{(z-a)^2} + \frac{z(z-a)}{(z-a)^2} $$ $$ = \frac{az + z^2 - az}{(z-a)^2} $$ $$ = \frac{z^2}{(z-a)^2} $$This result matches the given Z-transform $X(z)$.
Conclusion
Therefore, the causal signal corresponding to the Z-transform $X(z) = \frac{z^2}{(z-a)^2}$ is $(n+1) a^n u[n]$.
Checking the Options:
- Option 1: $(n + 1) a^n u[n]$. Its Z-transform is $\frac{z^2}{(z-a)^2}$. This is the correct match.
- Option 2: $a^{2n} u[n]$. Its Z-transform is $\frac{z}{z-a^2}$. Incorrect.
- Option 3: $n^{-1} a^n u[n]$. This signal is problematic for $n=0$. Its Z-transform is not standard. Incorrect.
- Option 4: $n^2 a^n u[n]$. Its Z-transform is $\frac{az(z+a)}{(z-a)^3}$. Incorrect.
The correct option is the first one.
Important Questions from Z Transform
The z transform of the following real exponential sequence
x(n) = {a n ;n >= 0} , {= 0 ; n < 0} and a > 0 is given by
The z transform of e −t sampled at 10 Hz will be:
Consider the z-transform X (z) = 5z2 +4z-1 + 3; 0 < |z| < ∞. The inverse z-transform x[n] is
The ROC of a system is the
The similarity between the Fourier transform and the z-transform is that
