View source for Signal Representation/The Convolution Theorem and Operation

 
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==Convolution in the time domain==
<br>
The&nbsp; &raquo;Convolution Theorem&laquo;&nbsp; is one of the most important laws of the Fourier transform,&nbsp; to which an own subchapter is dedicated in this tutorial.&nbsp; We will first consider the convolution theorem in the time domain and assume that the spectra of two time functions&nbsp; $x_1(t)$&nbsp; and&nbsp; $x_2(t)$&nbsp; are known:
 
:$$X_1 ( f )\hspace{0.15cm}\bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\hspace{0.15cm}x_1( t ),\quad X_2 ( f )\hspace{0.1cm}\bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\hspace{0.1cm}x_2 ( t ).$$

Then for the time function of the product&nbsp; $X_1(f) \cdot X_2(f)$&nbsp; applies:

:$$X_1 ( f ) \cdot X_2 ( f )\hspace{0.15cm}\bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\hspace{0.15cm}\int_{ - \infty }^{ + \infty } {x_1 ( \tau  )}  \cdot x_2 ( {t - \tau } )\hspace{0.1cm}{\rm d}\tau.$$
 
$\tau$&nbsp; is a formal integration variable with the dimension of a time.

{{BlaueBox|TEXT=  
$\text{Definition:}$&nbsp; 

The above connection of the time function&nbsp; $x_1(t)$&nbsp; and&nbsp; $x_2(t)$&nbsp; is called&nbsp; &raquo;'''convolution'''&laquo;.&nbsp; One represents this functional connection with a star:
 
:$$x_{\rm{1} } (t) * x_{\rm{2} } (t) = \int_{ - \infty }^{ + \infty } {x_1 ( \tau  ) }  \cdot x_2 ( {t - \tau } ) \hspace{0.1cm}{\rm d}\tau =  x_{\rm{2} } (t) * x_{\rm{1} } (t) .$$

*Thus the above Fourier correspondence can be written as follows:

:$$X_1 ( f ) \cdot X_2 ( f )\hspace{0.15cm}\bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\hspace{0.15cm}{ {x} }_{\rm{1} } ( t ) * { {x} }_{\rm{2} } (t ).$$

*The convolution is&nbsp; &raquo;commutative&laquo;  &nbsp; ⇒  &nbsp;The order of the operands can be changed:
 
:$$x_1(t) * x_2(t) =x_2(t) * x_1(t).$$

*The&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation#Proof_of_the_Convolution_Theorem|&raquo;Proof&laquo;]]&nbsp; will be shown at the end of this chapter.}}



{{GraueBox|TEXT=
[[File:EN_Sig_T_3_4_S1.png|right|frame|On calculation of signal and spectrum of  the LTI output]]  
$\text{Example 1:}$&nbsp; Every linear time-invariant&nbsp; $\rm (LTI)$&nbsp; system can be described both, 
#by the frequency response&nbsp; $H(f)$,
#by the impulse response&nbsp; $h(t)$. 


The relation between these two system quantities is also given by the Fourier transform.

*If the signal&nbsp; $x(t)$&nbsp; with spectrum&nbsp; $X(f)$&nbsp; is applied to the input,&nbsp; then the spectrum of the output signal is:
 
:$$Y(f) = X(f) \cdot H(f)\hspace{0.05cm}.$$

*It is possible to calculate the output signal in the time domain with the convolution theorem:
 
:$$y( t ) = x(t) * h( t ) = \int_{ - \infty }^{ + \infty } \hspace{-0.15cm}{x( \tau  )}  \cdot h( {t - \tau } )\hspace{0.1cm}{\rm d}\tau =  \int_{ - \infty }^{ + \infty } \hspace{-0.15cm} {h( \tau  )}  \cdot x( {t - \tau } )\hspace{0.1cm}{\rm d}\tau = h(t) * x( t ).$$

*This equation shows again the&nbsp; &raquo;commutativity&laquo;&nbsp; of the&nbsp; &raquo;convolution operation&laquo;.  }}


==Convolution in the frequency domain==
<br>
The duality between time and frequency domain also allows statements regarding the spectrum of the product signal:
 
:$$x_1 ( t ) \cdot x_2 ( t )\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,X_1 (f) * X_2 (f) =  \int_{ - \infty }^{ + \infty } {X_1 ( \nu  )}  \cdot X_2 ( {f - \nu })\hspace{0.1cm}{\rm d}\nu.$$

This result can be proved similarly to the&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation#Proof_of_the_Convolution_Theorem|&raquo;convolution in the time domain&laquo;]].&nbsp; However,&nbsp; the integration variable&nbsp; $\nu$&nbsp; now has the dimension of a frequency.

{{GraueBox|TEXT= 
[[File:EN_Sig_T_3_4_S2.png|right|frame|Convolution in the frequency domain&nbsp; $($e.g.&nbsp;  DSB&ndash;AM$)$]] 
$\text{Example 2:}$&nbsp; The &nbsp; [[Modulation_Methods/Zweiseitenband-Amplitudenmodulation#Beschreibung_im_Zeitbereich|
&raquo;Double-Sideband Amplitude Modulation&raquo;]]&nbsp; $\text{(DSB-AM)}$&nbsp; with carrier suppression is described by the drawn model.

*The time domain representation&nbsp; $\rm (blue)$&nbsp; shows the modulated signal&nbsp; $s(t)$&nbsp; as the product of the source signal&nbsp; $q(t)$&nbsp; and the&nbsp; $($normalized$)$&nbsp; carrier signal&nbsp; $z(t)$.


*According to the convolution theorem it follows for the frequency domain&nbsp; $\rm (red)$&nbsp; that the output spectrum&nbsp; $S(f)$&nbsp; is equal to the convolution product of&nbsp;  $Q(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \ q(t)$&nbsp; and&nbsp; $Z(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \ z(t)$.}}


==Convolution of a function with a Dirac delta function== 
<br>
The convolution operation becomes very simple,&nbsp; if one of the operands is a&nbsp; [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal#Dirac_.28delta.29_function_in_frequency_domain|&raquo;Dirac delta function&laquo;]].&nbsp; This applies both to the convolution in  time and frequency domain.

*We will consider the convolution of a function&nbsp; $x_1(t)$&nbsp; with the function
 
:$$x_2 ( t ) = \alpha  \cdot \delta ( {t - T} ) \quad \circ\,\!\!\!-\!\!\!-\!\!\!-\!\!\bullet \quad X_2 ( f )= \alpha \cdot  {\rm{e}}^{ - {\rm{j}}\hspace{0.05cm}\cdot\hspace{0.05cm}2\hspace{0.03cm}{\rm{\pi }}\hspace{0.05cm}\cdot\hspace{0.05cm}f\hspace{0.05cm}\cdot\hspace{0.05cm}T}.$$

*For the spectral function of the signal&nbsp; $y(t) = x_1(t) \ast x_2(t)$&nbsp; it follows:
 
:$$Y( f ) = X_1 ( f ) \cdot X_2 ( f ) = X_1 ( f ) \cdot  \alpha  \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.05cm}\cdot\hspace{0.05cm}2\hspace{0.03cm}{\rm{\pi }}\hspace{0.05cm}\cdot\hspace{0.05cm}f\hspace{0.05cm}\cdot\hspace{0.05cm}T}.$$

*The complex exponential function leads to a shift by&nbsp; $T$ &nbsp; &rArr; &nbsp; [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem|&raquo;Shifting Theorem&laquo;]],&nbsp; the factor&nbsp; $\alpha$&nbsp; to an attenuation&nbsp; $(\alpha < 1)$&nbsp; or amplification&nbsp; $(\alpha > 1)$.&nbsp; From this follows:
 
:$$x_1 (t) * x_2 (t) = \alpha  \cdot x_1 ( {t - T} ).$$

{{BlaueBox|TEXT=  
$\text{In Words: }$&nbsp; 
*The convolution of any function with a Dirac delta function at&nbsp; $t = T$&nbsp; results in the function shifted to the right by&nbsp; $T$,&nbsp;
 
*while the weighting of the Dirac delta function by the factor&nbsp; $\alpha$&nbsp; has to be taken into account.}}


{{GraueBox|TEXT=
$\text{Example 3:}$&nbsp; A rectangle pulse&nbsp; $x(t)$&nbsp; is delayed by an LTI system by the delay time&nbsp; $\tau = 3\,\text{ ms}$&nbsp; and attenuated by the factor&nbsp; $\alpha = 0.5$.
[[File:P_ID522__Sig_T_3_4_S3_neu.png|right|frame|Convolution of a rectangle pulse with a Dirac  delta  function]]  



Shift and attenuation can be recognised both 
*by the output signal&nbsp; $y(t)$,&nbsp;
 
*by the impulse response&nbsp; $h(t)$.
}}


==Graphical convolution==
<br>
[[File:P_ID2723__Sig_T_3_4_programm.png|right|frame|Screenshot of an older&nbsp; $($German$)$ version &nbsp;$\rm LNTwww$ applet&nbsp; &raquo;Convolution&laquo;:<br>&nbsp; &nbsp; $x_1(t)$&nbsp; is denoted as&nbsp; $x(t)$&nbsp; and&nbsp; $x_2(t)$&nbsp; as&nbsp; $h(t)$]]
For the descriptions in this section the following convolution operation is assumed:

:$$y(t) = x_1 (t) * x_2 (t) $$
:$$\Rightarrow \hspace{0.3cm}y(t) = \int_{ - \infty }^{ + \infty } {x_1 ( \tau  )}  \cdot x_2 ( {t - \tau } )\hspace{0.1cm}{\rm d}\tau.$$

The solution of the convolution integral shall be done graphically.&nbsp; It is assumed that&nbsp; $x_1(t)$&nbsp; and&nbsp; $x_2(t)$&nbsp; are continuous time signals. 

Then the following steps are required:
#&nbsp; &raquo;'''Change time variables'''&laquo;  of the functions:&nbsp;  $x_1(t) \to x_1(\tau)$, &nbsp; $x_2(t) \to x_2(\tau)$.
#&nbsp; &raquo;'''Mirroring the second function'''&laquo;: &nbsp; $x_2(\tau) \to x_2(-\tau)$.
#&nbsp; &raquo;'''Shifting the mirrorred function'''&laquo; by&nbsp; $t$:&nbsp;  $x_2(-\tau) \to x_2(t-\tau)$.
#&nbsp; &raquo;'''Multiplication'''&laquo;&nbsp; of both functions&nbsp; $x_1(\tau)$&nbsp; and&nbsp; $x_2(t-\tau)$.
#&nbsp; &raquo;'''Integration'''&laquo;&nbsp; over the product resp.&nbsp; $\tau$&nbsp; between the limits&nbsp; $-\infty$&nbsp; to&nbsp; $+\infty$.



<u>Note:</u>
*Since convolution is commutative,&nbsp; $x_1(\tau)$&nbsp; can also be mirrored instead of&nbsp; $x_2(\tau)$.

*The topic is also illustrated by the HTML5 applet&nbsp; [[Applets:Graphical_Convolution|&raquo;Graphical Convolution&laquo;]]. 
<br clear=all>
{{GraueBox|TEXT= 
[[File:P_ID582__Sig_T_3_4_S4_neu.png|right|frame|Jump function convoluted with the exponential function]] 
$\text{Example 4:}$&nbsp;

The procedure for the graphical convolution is now explained with a detailed example: 
#At the input of a filter there is a jump function&nbsp; $x(t) = \gamma(t)$. 
#The impulse response of the first order low-pass filter is&nbsp; $h( t ) = {1}/{T} \cdot {\rm{e} }^{ - t/T}.$
#The time axis is already renamed to&nbsp; $\tau$.


The graphic shows 
#the&nbsp; $($red colored$)$&nbsp;  input signal&nbsp; $x(\tau)$, 
#the&nbsp; $($blue colored$)$&nbsp; impulse response&nbsp; $h(\tau)$,&nbsp; 
#the&nbsp; $($grey colored$)$&nbsp; output signal&nbsp; $y(\tau)$. 


The output signal can be calculated for example using the following equation:
 
:$$y(t) = h(t) * x(t) = \int_{ - \infty }^{ + \infty } {h( \tau  )}  \cdot x( {t - \tau } )\hspace{0.1cm}{\rm d}\tau.$$

Some more remarks:
*The output value at&nbsp; $t = 0$&nbsp; is obtained by mirroring the input signal&nbsp; $x(\tau)$.&nbsp; This mirrored signal&nbsp; $x(-\tau)$&nbsp; is multiplied by the impulse response&nbsp; $h(\tau)$&nbsp; and integrated over it.

*As there is no time interval,&nbsp; where both the blue curve&nbsp; $h(\tau)$&nbsp; and at the same time also the red dashed mirrored curve&nbsp; $x(-\tau)$&nbsp; is not equal to zero,&nbsp; the result is&nbsp; $y(t=0)=0$.

*For any other time&nbsp; $t$&nbsp; the input signal must be shifted &nbsp; ⇒ &nbsp; $x(t-\tau)$,&nbsp; for example according to the green dotted curve for&nbsp; $t=T$.

*As in this example&nbsp; $x(t-\tau)$&nbsp; is only&nbsp; $0$&nbsp; or&nbsp; $1$&nbsp; the integration&nbsp; $($general from&nbsp; $\tau_1$&nbsp; to&nbsp; $\tau_2)$&nbsp; is very simple and you get here with&nbsp; $\tau_1 = 0$&nbsp;  and&nbsp; $\tau_2 = t$&nbsp;:
:$$y( t) = \int_0^{\hspace{0.05cm} t} {h( \tau)}\hspace{0.1cm} {\rm d}\tau = \frac{1}{T}\cdot\int_0^{\hspace{0.05cm} t} {{\rm{e}}^{ - \tau /T } }\hspace{0.1cm} {\rm d}\tau = 1 - {{\rm{e}}^{ - t /T } }.$$
*This sketch is valid for &nbsp; $t=T$&nbsp; and results in the output value&nbsp; $y(t=T) = 1 - 1/\text{e} \approx 0.632$.}}


==Descriptive interpretation of convolution==
<br>
We assume an impulse response&nbsp; $h(t)$&nbsp; which is first constant for one millisecond and then decreases linearly to zero until &nbsp; $t = 3 \,\text{ms}$.
[[File:P_ID524__Sig_T_3_4_S5_rah.png|right|frame|An interpretation of&nbsp; "Convolution"]] 
*If&nbsp; $K_0 \cdot \delta(t)$&nbsp; is applied to the input of this low-pass filter,&nbsp; the output signal&nbsp; $y(t)$&nbsp; has the same shape as the impulse response&nbsp; $h(t)$.&nbsp; The situation is shown in red in the figure.

*The&nbsp; $($blue$)$&nbsp; $T= 1 \,\text{ms}$&nbsp; shifted  Dirac delta with weight&nbsp; $K_1 > K_0$&nbsp; results in the output signal&nbsp; $y_1(t)$&nbsp; which is delayed with respect to the red signal and increased in amplitude.


Now we consider the input signal consisting of seven differently weighted Dirac deltas:
 
:$$x( t ) = \sum\limits_{n = 0}^6 {K_n  \cdot \delta ( {t - n \cdot T} ),}$$

which can be understood as a time&ndash;discrete approximation of a time&ndash;continuous signal. 

*The output signal of the linear system is the sum of the seven partial signals marked with different colors in the figure:
 
:$$y( t ) = \sum\limits_{n = 0}^6 {K_n  \cdot h( {t - n \cdot T} ).}$$

*We now look at the signal value at time&nbsp; $t = 4.5T$&nbsp; $($see dotted line$)$:
 
:$$y( {t = 4.5T} ) = K_2  \cdot h( {2.5T} ) + K_3  \cdot h(1.5 T ) + K_4  \cdot h( 0.5 T ).$$

*$y( {t = 4.5T} )$&nbsp; is thus only determined by the input signal values&nbsp; $K_2$,&nbsp; $K_3$&nbsp; and&nbsp; $K_4$.&nbsp; <u>Note:</u>

#The influence of&nbsp; $K_4$&nbsp; is  the greatest due to&nbsp; $h(0.5T) = 1$.&nbsp; 
#The influence of&nbsp; $K_3$&nbsp; is  less  strong due to&nbsp; $h(1.5T) = 0.75$.&nbsp; 
#The influence of&nbsp; $K_2$&nbsp; is the lowest due to&nbsp; $h(2.5T) = 0.25$.&nbsp;



==Proof of the Convolution Theorem==
<br>
{{BlaueBox|TEXT=  
$\text{Definition: }$&nbsp;
The following relation of time functions&nbsp; $x_1(t)$&nbsp; and&nbsp; $x_2(t)$&nbsp; is called the&nbsp; &raquo;'''convolution'''&laquo;&nbsp; and represents this functional relation with a star:
 
:$$x_{\rm{1} } (t) * x_{\rm{2} } (t) = \int_{ - \infty }^{ + \infty } {x_1 ( \tau  ) }  \cdot x_2 ( {t - \tau } ) \hspace{0.1cm}{\rm d}\tau.$$

*This results in the following&nbsp; &raquo;'''Fourier correspondence'''&laquo;:
 
:$$X_1 ( f ) \cdot X_2 ( f )\hspace{0.1cm}\bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\hspace{0.1cm}{ {x} }_{\rm{1} } ( t ) * { {x} }_{\rm{2} } (t ).$$}}


{{BlaueBox|TEXT=  
$\text{Proof: }$&nbsp;
The Fourier integrals of functions&nbsp; $x_1(t)$&nbsp; and&nbsp; $x_2(t)$&nbsp; are with modified integration variables:
 
:$$X_1 ( f ) = \int_{ - \infty }^{ + \infty } {x_1 ( \tau  )}  \cdot {\rm{e} }^{ - {\rm{j} }2{\rm{\pi } }f\tau }\hspace{0.1cm} {\rm{d } }\tau{\rm{,} }$$

:$$X_2 ( f ) = \int_{ - \infty }^{ + \infty } {x_2 ( {t\hspace{0.05cm}'} ) }  \cdot {\rm{e} }^{ - {\rm{j} }2{\rm{\pi } }ft\hspace{0.05cm}'}\hspace{0.1cm} {\rm{d} }t\hspace{0.05cm}'{\rm{.} }$$
 
*If you form the product of the spectral functions, you get
 
:$$X_1 (f) \cdot X_2 (f) = \int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {x_1 ( \tau  )  \hspace{0.05 cm}\cdot } }\hspace{0.05 cm} x_2 ( {t\hspace{0.05cm}'} ) \cdot {\rm{e} }^{ - {\rm{j} }2{\rm{\pi } }f\left( {\tau  + t\hspace{0.05cm}'} \right) }\hspace{0.1cm} {\rm d} \tau \hspace{0.1cm}{\rm d}t\hspace{0.05cm}'{\rm{.} }$$

*With the substitution&nbsp; $t = \tau + t\hspace{0.05cm}'$&nbsp; results:
 
:$$X_1 ( f ) \cdot X_2 ( f ) = \int_{ - \infty }^{ + \infty } {\left[ {\int_{ - \infty }^{ + \infty } {x_1 ( \tau )}  \cdot x_2 ( {t - \tau} )\hspace{0.1cm}{\rm{d } } }\tau \right] }  \cdot {\rm{e} }^{ - {\rm{j} }2{\rm{\pi } }ft}\hspace{0.1cm} {\rm{d} }t{\rm{.} }$$

:This equation already takes into account that the exponential function is independent of the inner integration variable &nbsp;  &rArr; &nbsp; $τ$&nbsp; acts only as factor of the inner integral.

*If we now denote the product of the two spectra with&nbsp; $P(f)$&nbsp; and the corresponding time function with&nbsp; $p(t)$, the corresponding Fourier integral is
 
:$$P(f) = X_1 ( f ) \cdot X_2 ( f ) =\int_{ - \infty }^{ + \infty } {p( t )}  \cdot {\rm{e} }^{ - {\rm{j} }2{\rm{\pi } }ft} \hspace{0.1cm}{\rm{d} }t{\rm{.} }$$

*A coefficient comparison of the last two integrals shows that the following relationship apply:
 
:$$p( t ) = \int_{ - \infty }^{ + \infty } {x_1 ( \tau  )}  \cdot x_2 ( {t - \tau } )\hspace{0.1cm}{\rm{d } }\tau{\rm{.} }$$

<div align="right">q.e.d.</div>}}


==Exercises for the chapter==
<br>
[[Aufgaben:Exercise_3.7:_Synchronous_Demodulator|Exercise 3.7: Synchronous Demodulator]]

[[Aufgaben:Exercise_3.7Z:_Rectangular_Signal_with_Echo|Exercise 3.7Z: Rectangular Signal with Echo]]

[[Aufgaben:Exercise_3.8:_Triple_Convolution|Exercise 3.8: Triple Convolution?]]

[[Aufgaben:Exercise_3.8Z:Convolution_of_Two_Rectangles|Exercise 3.8Z: Convolution of Two Rectangles]]

[[Aufgaben:Exercise_3.9:_Convolution_of_Rectangle_and_Gaussian_Pulse|Exercise 3.9: Convolution of Rectangle and Gaussian Pulse]]

[[Aufgaben:Exercise_3.9Z:_Convolution_of_Gaussian_Pulses|Exercise 3.9Z: Convolution of Gaussian Pulses]]



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