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
What is the set of all values of z for which X(z) attains a finite value?
The z transform of the following real exponential sequence
x(n) = {a n ;n >= 0} , {= 0 ; n < 0} and a > 0 is given by
What will be the z-transform of a Unit step function ?
The z-transform of a causal periodic signal can be determined from the knowledge of the z-transform of its:
Two discrete-time linear time-invariant systems with impulse responses
h 1[n] = δ[n - 1] + δ[n + 1] and h 2[n] = δ[n] + δ[n - 1] are connected in cascade, where δ[n] is the Kronecker delta. The impulse response of the cascaded system is
