The z transform of e −t sampled at 10 Hz will be:
Understanding the Z-Transform of a Sampled Signal
The question asks us to find the z-transform of the continuous-time signal \(e^{-t}\) after it has been sampled at a frequency of 10 Hz. To solve this, we first need to convert the continuous-time signal into a discrete-time signal by applying the sampling process, and then find the z-transform of the resulting discrete-time signal.
Steps to Calculate the Z-Transform of the Sampled Signal
Here are the steps we will follow:
- Identify the continuous-time signal \(x(t)\).
- Determine the sampling frequency (\(f_s\)) and calculate the sampling period (\(T\)).
- Obtain the discrete-time signal \(x[n]\) by sampling \(x(t)\) at \(t = nT\).
- Compute the z-transform \(X(z)\) of the discrete-time signal \(x[n]\).
Step 1: Identify the Continuous-Time Signal
The given continuous-time signal is:
\(x(t) = e^{-t}\)
Step 2: Determine Sampling Parameters
The sampling frequency is given as:
\(f_s = 10\) Hz
The sampling period \(T\) is the time interval between consecutive samples. It is calculated as the reciprocal of the sampling frequency:
\(T = \frac{1}{f_s} = \frac{1}{10} = 0.1\) seconds
Step 3: Find the Sampled Discrete-Time Signal
The discrete-time signal \(x[n]\) is obtained by evaluating the continuous-time signal \(x(t)\) at integer multiples of the sampling period \(T\). So, \(t = nT\).
\(x[n] = x(nT) = e^{-(nT)}\)
Substitute the value of \(T = 0.1\) seconds into the expression:
\(x[n] = e^{-(0.1)n}\)
This discrete-time signal can also be written as:
\(x[n] = (e^{-0.1})^n\)
This is the sequence for which we need to find the z-transform.
Step 4: Compute the Z-Transform
The z-transform of a discrete-time signal \(x[n]\) is defined as:
\(X(z) = \sum_{n=0}^{\infty} x[n] z^{-n}\)
Substitute the expression for \(x[n]\) that we found:
\(X(z) = \sum_{n=0}^{\infty} (e^{-0.1})^n z^{-n}\)
This summation can be rewritten as:
\(X(z) = \sum_{n=0}^{\infty} (e^{-0.1} z^{-1})^n\)
This is an infinite geometric series of the form \(\sum_{n=0}^{\infty} r^n\), where the common ratio \(r = e^{-0.1} z^{-1}\). The sum of an infinite geometric series is given by \(\frac{1}{1-r}\), provided that \(|r| < 1\).
Using the formula for the sum of the geometric series:
\(X(z) = \frac{1}{1 - e^{-0.1} z^{-1}}\)
To get the z-transform in a more standard form (with positive powers of \(z\)), we multiply the numerator and the denominator by \(z\):
\(X(z) = \frac{1 \cdot z}{(1 - e^{-0.1} z^{-1}) \cdot z} = \frac{z}{z - e^{-0.1}}\)
Now, we need to evaluate the value of \(e^{-0.1}\). Using a calculator, we find:
\(e^{-0.1} \approx 0.904837\)
Rounding this value to one decimal place, we get approximately 0.9.
So, the z-transform is approximately:
\(X(z) \approx \frac{z}{z - 0.9}\)
Comparing this result with the given options:
| Option | Expression | Comparison |
|---|---|---|
| 1 | \(\frac{z}{z−10}\) | \(\frac{z}{z - e^{-0.1}} \neq \frac{z}{z-10}\) |
| 2 | \(\frac{z}{z−0.1}\) | \(\frac{z}{z - e^{-0.1}} \neq \frac{z}{z-0.1}\) |
| 3 | \(\frac{z}{z−0.9}\) | \(\frac{z}{z - e^{-0.1}} \approx \frac{z}{z - 0.9}\) (since \(e^{-0.1} \approx 0.9\)) |
| 4 | \(\frac{z}{z−1.1}\) | \(\frac{z}{z - e^{-0.1}} \neq \frac{z}{z-1.1}\) |
The calculated z-transform \(\frac{z}{z - e^{-0.1}}\) closely matches option 3 when \(e^{-0.1}\) is approximated as 0.9.
Revision Table: Z-Transform Concepts
| Concept | Definition/Formula | Relevance to Problem |
|---|---|---|
| Sampling Frequency (\(f_s\)) | Rate at which a continuous signal is sampled (Hz). | Given as 10 Hz. |
| Sampling Period (\(T\)) | Time between samples, \(T = 1/f_s\). | Calculated as 0.1 seconds. |
| Sampled Signal (\(x[n]\)) | Discrete-time signal obtained as \(x(nT)\). | Found as \(e^{-0.1n}\). |
| Z-Transform (\(X(z)\)) | Summation \(\sum_{n=0}^{\infty} x[n] z^{-n}\) (unilateral). | Applied to \(x[n]\) to get \(X(z)\). |
| Geometric Series Sum | \(\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}\) for \(|r| < 1\). | Used to evaluate the Z-transform sum. |
| Z-Transform of \(a^n u[n]\) | Standard transform pair: \(\frac{z}{z-a}\). | \(x[n] = (e^{-0.1})^n u[n]\) implies \(a = e^{-0.1}\). |
Additional Information on Sampled Signal Z-Transforms
When you sample a continuous-time signal \(x(t)\) with a sampling period \(T\), the resulting discrete-time signal is \(x[n] = x(nT)\). Finding the z-transform of this discrete signal is a common operation in digital signal processing.
A standard z-transform pair is that the z-transform of the discrete-time signal \(a^n u[n]\) is \(\frac{z}{z-a}\), where \(u[n]\) is the unit step function (which is 1 for \(n \ge 0\) and 0 for \(n < 0\)). Since the summation for the unilateral z-transform starts from \(n=0\), we usually assume the signal begins at \(n=0\), effectively multiplying by \(u[n]\).
In this problem, the sampled signal is \(x[n] = e^{-0.1n} = (e^{-0.1})^n\). This directly matches the form \(a^n\) with \(a = e^{-0.1}\). Therefore, the z-transform is \(\frac{z}{z - e^{-0.1}}\).
The value of \(e^{-0.1}\) is approximately 0.9048. Options are often provided with rounded values in multiple-choice questions. The option \(\frac{z}{z-0.9}\) uses a rounded value of \(e^{-0.1}\) to one decimal place. This is a common practice in exam questions where exact values might lead to complicated options.
Important Questions from Z Transform
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: For a given sequence $X[n]=(\frac{1}{2})^n u(n)$ and its Z-transform $X(z)=\frac{1}{1-\frac{1}{2}z^{-1}}, \quad |z|>\frac{1}{2}$ is a stable system.
Reason R: For a stable system ROC (Region of Convergence) must include the unit circle.
In the light of the above statements, choose the most appropriate answer from the options given below
Match the following :
List - I List - II (a) U(t) (i) \(\dfrac{TZ}{(Z-1)^{2}}\) (b) t (ii) \(\dfrac{Z}{Z-e^{-aT}}\) (c) t2 (iii) \(\dfrac{Z}{Z-1}\) (d) e-at (iv) \(\dfrac{T^{2}Z(Z+1)}{(Z-1)^{3}}\) Codes :
For the following signal :
\(x(n)=\left(\dfrac{1}{2}\right)^{n}u(n)+\left(\dfrac{1}{3}\right)^{n}u(n)\)
Z transforms and ROC has been given in below statements :
(a) \(Z[x(n)]=\dfrac{Z}{Z-\frac{1}{2}}+\dfrac{Z}{Z-\frac{1}{3}}\)
(b) ROC : \(|Z| \gt \dfrac{1}{2}\) and \(|Z| \gt \dfrac{1}{3}\)
(c) \(Z[x(n)]=\dfrac{Z}{Z+\frac{1}{2}}+\dfrac{Z}{Z+\frac{1}{3}}\)
(d) ROC : \(|Z| \lt \dfrac{1}{3}\) and \(|Z| \lt \dfrac{1}{3}\)
Out of the above given statements which are correct ?
Consider a discrete-time sequencebr
$x[n] = \begin{cases} (0.2)^n, & 0 \le n \le 7 \\ 0, & \text{otherwise} \end{cases}$
The region of convergence of $X(z)$, the z-transform of $x[n]$, consists ofFor a causal discrete-time LTI system with transfer function
$H(z) = \frac{2z^2+3}{(z+\frac{1}{3})(z-\frac{1}{3})}$,
which of the following statements is/are true?
