Continuous-time vs discrete-time signals in Signal Processing - Visual Side-by-Side Comparison

Concept Flow - Continuous-time vs discrete-time signals

Start: Signal Defined

↓

Is signal continuous?

Yes→Continuous-time signal

No↓

Discrete-time signal

↓

Sampled values at discrete times

↓

End

The flow shows deciding if a signal is continuous or discrete, then processing accordingly.

Execution Sample

Signal Processing

import numpy as np
import matplotlib.pyplot as plt

t = np.linspace(0, 1, 1000)  # continuous time
x_cont = np.sin(2 * np.pi * 5 * t)

n = np.arange(0, 11)  # discrete time
x_disc = np.sin(2 * np.pi * 5 * n / 10)

This code creates a continuous-time sine wave and a discrete-time sine wave sampled at 11 points.

Execution Table

Step	Variable	Value	Description
1	t	array of 1000 points from 0 to 1	Continuous time vector created
2	x_cont	sin(2π5t) values for each t	Continuous-time sine wave computed
3	n	array [0,1,...,10]	Discrete time indices created
4	x_disc	sin(2π5n/10) values for each n	Discrete-time sine wave computed
5	Plot	Continuous and discrete signals	Visual comparison of signals
6	End	-	Execution complete

💡 All steps executed to create and compare continuous and discrete signals

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	Final
t	undefined	array of 1000 points from 0 to 1	same	same	same	same
x_cont	undefined	undefined	array of sine values	same	same	same
n	undefined	undefined	undefined	array [0..10]	same	same
x_disc	undefined	undefined	undefined	undefined	array of sine values	same

Key Moments - 3 Insights

Why does the continuous-time signal have 1000 points but the discrete-time signal only 11?

The continuous-time signal is sampled very finely to simulate continuity (step 1), while the discrete-time signal uses specific integer time points (step 3). This difference is shown in the execution_table rows 1 and 3.

Why do both signals use the sine function but look different?

The continuous signal uses many points to form a smooth curve (step 2), while the discrete signal uses fewer points sampled at intervals (step 4). This causes the discrete signal to appear as dots or steps.

Can discrete-time signals represent continuous signals perfectly?

No, discrete signals approximate continuous signals by sampling at intervals. The finer the sampling (more points), the closer the approximation, but some information is always lost.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what does variable 'n' represent?

A Continuous time points

B Discrete time indices

C Signal amplitude values

D Frequency of the sine wave

💡 Hint

Check execution_table row 3 description for 'n'

At which step is the continuous-time sine wave calculated?

A Step 3

B Step 1

C Step 2

D Step 4

💡 Hint

Look at execution_table row 2 for 'x_cont'

If we increase the number of points in 'n' from 11 to 1000, what happens to the discrete signal?

A It becomes more like the continuous signal

B It becomes less smooth

C It disappears

D Frequency changes

💡 Hint

Refer to variable_tracker for how sample points affect signal shape

Concept Snapshot

Continuous-time signals are defined for every instant (smooth curve).
Discrete-time signals are defined only at specific time points (samples).
Discrete signals approximate continuous signals by sampling.
More samples mean better approximation.
Use numpy linspace for continuous and arange for discrete time vectors.

Full Transcript

This visual execution shows how continuous-time and discrete-time signals are created and compared. First, a continuous time vector 't' with 1000 points is made, then a sine wave 'x_cont' is computed for all these points, forming a smooth curve. Next, a discrete time vector 'n' with 11 integer points is created, and a discrete sine wave 'x_disc' is computed only at these points, appearing as separate dots. The key difference is that continuous signals have values at every instant, while discrete signals have values only at specific times. Sampling more points in discrete signals makes them closer to continuous signals. This step-by-step trace helps understand how signals differ and relate.

Practice

1. Which of the following best describes a continuous-time signal?

easy

A. A signal defined only at specific time points

B. A signal that is always zero

C. A signal that only takes integer values

D. A signal defined for every instant of time

Solution

1. Step 1: Understand continuous-time signal definition

   A continuous-time signal is defined for every instant in time, meaning it varies smoothly without gaps.

2. Step 2: Compare with discrete-time signals

   Discrete-time signals are defined only at specific time points, not continuously.

3. Final Answer:

   A signal defined for every instant of time -> Option D

4. Quick Check:

   Continuous-time = defined at all times [OK]

Hint: Continuous means all time points included [OK]

Common Mistakes:

• Confusing continuous with discrete signals
• Thinking discrete signals are continuous
• Assuming continuous signals only take integer values

2. Which syntax correctly represents a discrete-time signal sampled from a continuous signal x(t) at integer times n?

easy

A. x[n] = integral of x(t)

B. x(n) = x(t) for all t

C. x[n] = x(t) at t = n

D. x(t) = x[n] for n in integers

Solution

1. Step 1: Identify discrete-time signal notation

   Discrete-time signals use square brackets [n] to denote samples at integer times.

2. Step 2: Match sampling definition

   Sampling means taking values of x(t) at specific times t = n, so x[n] = x(t) at t = n is correct.

3. Final Answer:

   x[n] = x(t) at t = n -> Option C

4. Quick Check:

   Discrete-time uses square brackets and samples at integers [OK]

Hint: Discrete signals use square brackets for indexing [OK]

Common Mistakes:

• Using parentheses instead of square brackets for discrete signals
• Confusing continuous and discrete notation
• Thinking sampling is an integral operation

3. Given the continuous-time signal x(t) = 2t, what is the discrete-time signal x[n] sampled at integer times?

medium

A. x[n] = 2n

B. x[n] = 2t

C. x[n] = n/2

D. x[n] = 2

Solution

1. Step 1: Understand sampling at integer times

   Sampling means replacing t by integer n in x(t).

2. Step 2: Substitute t = n in x(t) = 2t

   We get x[n] = 2n.

3. Final Answer:

   x[n] = 2n -> Option A

4. Quick Check:

   Replace t with n in formula [OK]

Hint: Replace t with n for discrete samples [OK]

Common Mistakes:

• Leaving t as variable in discrete signal
• Dividing instead of multiplying by 2
• Confusing n and t variables

4. Identify the error in this discrete-time signal definition: x[n] = x(t) at t = n + 0.5

medium

A. The notation x[n] is incorrect

B. Sampling at non-integer times is invalid for discrete-time signals

C. The signal should be continuous, not discrete

D. There is no error; this is a valid discrete-time signal

Solution

1. Step 1: Understand discrete-time signal sampling points

   Discrete-time signals are defined at integer time points n.

2. Step 2: Check sampling time t = n + 0.5

   Sampling at half-integers is not standard discrete-time sampling; it breaks the integer indexing rule.

3. Final Answer:

   Sampling at non-integer times is invalid for discrete-time signals -> Option B

4. Quick Check:

   Discrete-time = integer time samples only [OK]

Hint: Discrete-time signals sample only integer times [OK]

Common Mistakes:

• Thinking discrete signals can sample at any time
• Confusing notation with error
• Assuming half-integers are valid discrete points

5. You have a continuous-time signal x(t) = sin(2πt). You sample it at 1 Hz (once per second). Which discrete-time signal x[n] do you get, and what is a potential problem?

hard

A. x[n] = sin(2πn); signal appears constant at zero

B. x[n] = sin(2πn); signal is perfectly captured

C. x[n] = sin(πn); signal is undersampled

D. x[n] = sin(4πn); signal is oversampled

Solution

1. Step 1: Write discrete-time signal from sampling

   Sampling at 1 Hz means t = n seconds, so x[n] = sin(2πn).

2. Step 2: Evaluate sin(2πn) for integer n

   Since sin(2πn) = 0 for all integers n, the sampled signal is always zero.

3. Step 3: Identify the problem

   This is an example of aliasing where the sampled signal loses information and appears constant.

4. Final Answer:

   x[n] = sin(2πn); signal appears constant at zero -> Option A

5. Quick Check:

   Sampling frequency equals signal frequency causes zero samples [OK]

Hint: Sampling at signal period causes zero values [OK]

Common Mistakes:

• Assuming sampling always captures signal perfectly
• Ignoring aliasing effects
• Confusing frequency and sampling rate