Math Problem Statement

Représenter graphiquement le résultat des convolutions des signaux x(t) et h(t) suivants (de a- à e-).

Solution

The problem you're working on asks to graphically represent the results of the convolutions of the signals x(t)x(t) and h(t)h(t), which are given in different forms (from aa to ee).

Explanation of Convolution:

Convolution is an operation that combines two signals x(t)x(t) and h(t)h(t) into a new signal, usually denoted as y(t)=(xh)(t)y(t) = (x * h)(t). The convolution operation is expressed as:

y(t)=x(τ)h(tτ)dτy(t) = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) d\tau

In simple terms, this operation "slides" one function across the other and integrates their product over all time shifts.


Interpretation of the Signals:

Case a:

  • h(t)h(t) is a triangular function centered between T0-T_0 and T0T_0, with a peak at t=0t = 0.
  • x(t)x(t) is a pair of Dirac delta functions located at t=T0t = -T_0 and t=T0t = T_0.

In this case, the convolution will result in a shifted triangular waveform, as the delta functions act like an "impulse" that samples h(t)h(t) at their locations.

Case b:

  • h(t)h(t) is the same triangular function as in Case a.
  • x(t)x(t) consists of two delta functions located at t=T0/2t = -T_0/2 and t=T0/2t = T_0/2.

Here, the convolution result will be two shifted versions of the triangular waveform, one centered around t=T0/2t = -T_0/2 and the other around t=T0/2t = T_0/2.

Case c:

  • h(t)h(t) is triangular but skewed (steeper on the left side).
  • x(t)x(t) remains the same as in Case b.

The convolution will result in a skewed triangular waveform that is shifted twice as in Case b, with slightly different shapes due to the skew in h(t)h(t).

Case d:

  • h(t)h(t) is a ramp function increasing from 0 to T0T_0.
  • x(t)x(t) is a single delta function at t=T0t = -T_0.

Since x(t)x(t) is a delta function, the convolution will result in a shifted ramp function.

Case e:

  • h(t)h(t) is similar to Case d, a ramp function.
  • x(t)x(t) is a pair of delta functions at t=T0/4t = -T_0/4 and t=T0t = T_0.

The convolution will yield two shifted versions of the ramp function, each shifted to correspond to the locations of the delta functions.


Next Steps

Would you like me to compute the exact equations for these convolutions or assist in graphing them for clarity?

Here are some questions for further understanding:

  1. What are the specific properties of delta functions in convolution?
  2. How do the shapes of signals affect the convolution result?
  3. What happens when the convolution involves more complex signals than simple triangles or ramps?
  4. How can the convolution operation be applied in practical engineering contexts?
  5. Can you think of any real-world systems where these types of convolutions occur?

Tip: Convolution with delta functions is a powerful tool since delta functions pick specific points from the signal, essentially "sampling" the other function.

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Math Problem Analysis

Mathematical Concepts

Convolution
Signal Processing
Delta Functions
Triangular and Ramp Functions

Formulas

y(t) = (x * h)(t)
y(t) = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) d\tau

Theorems

Convolution Theorem
Properties of Delta Functions

Suitable Grade Level

Undergraduate Engineering (Signal Processing)