In this section we will discuss the main properties of the FTC.
Screencast video [⯈]
Linearity
As a result of the fact that the integration operation of the FTC is a linear operation it follows that the FTC is linear.
\begin{eqnarray*}
x(t) \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ X(f) \quad\text{ and }\quad y(t) \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ Y(f)
\end{eqnarray*}
$$
\boxed{
\begin{eqnarray*}
a \cdot x(t) + b \cdot y(t) & \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ &
a \cdot X(f) + b \cdot Y(f)
\end{eqnarray*}}
$$
Proof of linearity property
\begin{eqnarray*}
\mathcal{F}\{a \cdot x(t) + b \cdot y(t)\} &&=\int_{-\infty}^{\infty} \left ( a \cdot x(t) + b \cdot y(t) \right ) e^{-j2 \pi f t} \text{d} t \\
&&= \int_{-\infty}^{\infty} \left ( a \cdot x(t) \right ) e^{-j2 \pi f t} \text{d} t +
\int_{-\infty}^{\infty} \left ( b \cdot y(t) \right ) e^{-j2 \pi f t} \text{d} t \\
&&= a \cdot \int_{-\infty}^{\infty} x(t) e^{-j2 \pi f t} \text{d} t +
b \cdot \int_{-\infty}^{\infty} y(t) e^{-j2 \pi f t} \text{d} t \\
&&= a \cdot X(f) + b \cdot Y(f)
\end{eqnarray*}
The linearity property implies that of the sum of signals equals the sum of the FTC's of the signals.
Example
As an example of this linear property we calculate the FTC of a sinusoidal signal.
By using Euler we can write the signal into two phasors.
Each of these phasors transforms into a delta pulse in frequency domain.
\begin{eqnarray*}
\text{Euler } \rightarrow \text{ }\cos(2 \pi f_0 t) &=& \frac{1}{2} e^{j2 \pi f_0 t}+\frac{1}{2} e^{-j2 \pi f_0 t} \\
e^{j2 \pi f_0 t} &\circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ& \delta(f-f_0) \\
e^{-j2 \pi f_0 t} &\circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ& \delta(f+f_0)
\end{eqnarray*}
By using the linearity property we can now write the FTC of the sum of phasors as the sum of the FTC's of the two individual phasors.
\begin{eqnarray*}
\Rightarrow \text{ } \mathcal{F}\{\cos(2 \pi f_0 t)\} &=& \mathcal{F}\{\frac{1}{2} e^{j2 \pi f_0 t}+\frac{1}{2} e^{-j2 \pi f_0 t}\}\\
&=& \frac{1}{2} \mathcal{F}\{e^{j2 \pi f_0 t}\}+\frac{1}{2} \mathcal{F}\{e^{-j2 \pi f_0 t}\}\\
&=& \frac{1}{2}\delta(f-f_0) + \frac{1}{2}\delta(f+f_0)
\end{eqnarray*}
Thus the FTC of a cosine signal consists of a sum of two delta pulses in the frequency domain.
The first step of the proof of the conjugation property is to move the conjugation sign outside the integral. Inside the brackets this results into an FTC from which the phasor has a positive sign, which writes as $X(-f)$.
\begin{eqnarray*}
y(t) = x^\ast(t) \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ
Y(f)&=&\int_{-\infty}^{\infty} x^\ast(t) e^{-j2 \pi f t} \text{d} t
= \left ( \int_{-\infty}^{\infty} x(t) e^{j2 \pi f t} \text{d} t \right )^* \\
&=& \left ( X(-f) \right )^\ast=X^\ast(-f)
\end{eqnarray*}
The final notation $X^\ast(-f)$ is just simpler way of the notation $(X(-f))^\ast$.
An important result of this conjugation property is that FTC $X(f)$ of a real signal $x(t)$, thus when $x(t)=x^\ast(t)$, has the following symmetry properties:
The magnitude response is symmetric, while the phase response is anti-symmetric, thus:
$$
x(t) \text{ real } \Rightarrow \hspace{2mm} x(t)=x^\ast(t) \hspace{2mm} \Rightarrow \hspace{2mm} X(f)=X^\ast(-f) \hspace{2mm} \Rightarrow
$$
\begin{eqnarray*}
|X(f)| = |X(-f)| &:& \quad\textbf{Magnitude response symmetric}\\
\angle \{X(f)\} = - \angle \{X(f)\} &:& \quad\textbf{Phase response anti-symmetric}
\end{eqnarray*}
Example
A simple example of these symmetry properties can be found in the FTC of a shifted delta pulse. For this case the time domain signal is a real and the FTC results in a complex exponent as follows:
\begin{eqnarray*}
x(t)=\delta(t-t_0) &\circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ X(f)&= \int_{-\infty}^{\infty} x(t) e^{-j2 \pi f t} \text{d} t \\
&&= \int_{-\infty}^{\infty} \delta(t-t_0) e^{-j2 \pi f t} \text{d} t
= e^{-j2 \pi f t_0}
\end{eqnarray*}
From this result it follows that the magnitude response equals 1, while the phase response equals $-2\pi t_0$.
\begin{eqnarray*}
\textbf{Symmetric Magnitude response} &:& \quad |X(f)|=|e^{-j2 \pi f t_0}|=1 \\
\textbf{Anti-symmetric Phase response} &:& \quad \angle \{X(f)\} = -2 \pi t_0
\end{eqnarray*}
A plot of this magnitude- and phase response is depicted in the figure below.
Magnitude and phase response of shifted delta pulse.
From this plot the symmetry properties are clearly visible.
Note finally that for this case the angle of the phase response is directly related to the shift in time domain.
$$
y(t) = x(k \cdot t) \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ Y(f) = \int_{-\infty}^{\infty} x(k \cdot t) e^{-j2 \pi f t}
$$
For the proof of the scaling property it is the easiest to split the proof into two cases.
For $k>0$ we substitute for the scaled variable $k \cdot t$ a new variable $\tau$ which leads straightforward to the following result:
$$
Y(f) \overset{\tau =k \cdot t}{=} \int_{-\infty}^{\infty} x(\tau) e^{-j2\pi f \frac{1}{k} \tau} \cdot \frac{1}{k} \text{d} \tau
= \frac{1}{k} X(\frac{f}{k})
$$
For $k<0$ we follow the same procedure. But now the boundaries of the integral are changed. Changing these boundaries causes an extra minus sign. As a last step we take the absolute value of $k$ which combines the negative sign together with the negative value of $k$.
\begin{eqnarray*}
Y(f) &\overset{\tau =k \cdot t}{=}& \int_{\color{red}{+\infty}}^{\color{red}{-\infty}} x(\tau) e^{-j2\pi f \frac{1}{k} \tau} \cdot \frac{1}{k} \text{d} \tau \\
&=& \frac{\color{red}{-1}}{k} \int_{\color{red}{-\infty}}^{\color{red}{+\infty}} x(\tau) e^{-j2\pi f \frac{1}{k} \tau} \text{d} \tau = \frac{1}{|k|} X\left(\frac{f}{k}\right)
\end{eqnarray*}
Example
previously we have shown that the FTC of a block signal results in a sinc function, as given in the following equation and depicted in the figure below.
$$
x(t)= \begin{cases}
1 & \text{for } -\frac{T_0}{2} < t < \frac{T_0}{2} \\
0 & \text{elsewhere}
\end{cases}
\hspace{3mm} \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ \hspace{3mm}
X(f) = T_0 \frac{\sin (\pi f T_0)}{\pi f T_0}
$$
Result of scale operation applied to a block function.
The figure shows an example in which applied a scale factor of 2: The time domain signal has shrunk by a factor of 2, while the frequency domain signal is stretched by a factor 2.
This leads to the following important "rule of thumb":
Shrinking in one domain implies stretching in the other domain.
The first step of the proof is to change everywhere the variable $t$ by $t-t_0$.
\begin{eqnarray*}
x(t-t_0) &\circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ & \int_{-\infty}^{\infty} x(t-t_0) e^{-j2 \pi f t} \text{d} t \\
&=& \int_{-\infty}^{\infty} x(\color{red}{t-t_0}) e^{-j2 \pi f (\color{red}{t-t_0}+ \color{blue}{t_0})} \text{d} (\color{red}{t-t_0})
\end{eqnarray*}
In the next step we take out the negative exponent with the term $t_0$ and substitute $t-t_0$ by $\tau$, which leads to the final result as follows:
\begin{eqnarray*}
&\overset{\color{red}{\tau}=t-t_0}{=}& e^{-j2 \pi f \color{blue}{t_0}} \cdot \int_{-\infty}^{\infty} x(\color{red}{\tau}) e^{-j2 \pi f \color{red}{\tau}} \text{d} {\color{red}\tau} \\
&=& e^{-j2 \pi f t_0} \cdot X(f)
\end{eqnarray*}
The steps of this proof are similar to the previous ones and are as follows:
\begin{eqnarray*}
\mathcal{F}^{-1}\{X(f-f_0)\} &=& \int_{-\infty}^{\infty} X(f-f_0) e^{j2 \pi f t} \text{d} f \\
&=& \int_{-\infty}^{\infty} X({\color{red}f-f_0}) e^{j2 \pi ({\color{red}f-f_0} + {\color{blue}f_0}) t} \text{d} ({\color{red}f-f_0}) \\
&\overset{{\color{red}\eta}=f-f_0}{=}& e^{j2 \pi {\color{blue}f_0} t} \cdot \int_{-\infty}^{\infty} X({\color{red}\eta}) e^{j2 \pi {\color{red}\eta} t} \text{d} {\color{red}\eta}
= e^{j2 \pi f_0 t} \cdot x(t)
\end{eqnarray*}
Modulation
A result of this frequency shift property is the following modulation property:
\begin{eqnarray*}
e^{j2 \pi f_0 t} \cdot x(t) + e^{-j2 \pi f_0 t} \cdot x(t)
&\circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ& X(f-f_0) + X(f+f_0) \newline
\Leftrightarrow \hspace{3mm} 2 {\color{red}\cos (2 \pi f_0 t) \cdot } x(t) & \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ & X(f-f_0) + X(f+f_0)
\end{eqnarray*}
Thus when multiplying, or modulating, a signal $x(t)$ with a sinusoidal signal, the result in the frequency domain is a sum of two shifted versions of the FTC $X(f)$ of signal $x(t)$ which results into the following “rule of thumb”:
Modulation in one domain is shifting in the other domain..
The first step of the proof of the time domain convolution property is to use the definition of the convolution integral into the FTC definition.
\begin{eqnarray*}
\mathcal{F}\{\color{red}{x(t) \star y(t)}\} &=& \int_{t=-\infty}^{\infty} \left ( {\color{red}\int_{\tau= -\infty}^{\infty} x(\tau) y(t-\tau) \text{d} \tau }\right ) e^{-j2 \pi f t} \text{d} t
\end{eqnarray*}
Then by interchanging the order of the integrals and introducing in the second integral everywhere the variable $t-\tau$, this results in an extra phasor component $e^{-j2 \pi f \tau}$ in the first integral.
\begin{eqnarray*}
\ldots
&=& \int_{\tau=-\infty}^{\infty} x(\tau) \left ( \int_{t=-\infty}^{\infty} y(t-\tau)
e^{-j2 \pi f t} \text{d} t \right ) \text{d} \tau \\
&=& \int_{\tau=-\infty}^{\infty} x(\tau) e^{-j2 \pi f {\color{red}\tau}} \left ( \int_{t=-\infty}^{\infty} y(t-\tau) e^{-j2 \pi f (t-{\color{red}\tau}) } \text{d} t \right ) \text{d} \tau
\end{eqnarray*}
Finally by substituting $t-\tau$ by a new variable $p$ we obtain the final result.
\begin{eqnarray*}
\ldots &=& \left ( \int_{\tau=-\infty}^{\infty} x(\tau) e^{-j2 \pi f \tau} \text{d} \tau \right ) \cdot \left ( \int_{p=-\infty}^{\infty} y(p) e^{-j2 \pi f p } \text{d} p \right ) \\
&=& X(f) \cdot Y(f)
\end{eqnarray*}
The first step of the proof of the frequency domain convolution property is to use the definition of Inverse FTC for the signal $x(t)$.
\begin{eqnarray*}
\mathcal{F}\{x(t) \cdot y(t)\} &=& \int_{t=-\infty}^{\infty} \color{red}{x(t)} \cdot y(t) e^{-j2 \pi f t} \text{d} t\\
&=& \int_{t=-\infty}^{\infty} \left ( \color{red}
{\int_{\eta=-\infty}^{\infty} X(\eta) e^{j2 \pi \eta t} \text{d} \eta }
\right ) \cdot y(t) e^{-j2 \pi f t} \text{d} t
\end{eqnarray*}
Then by interchanging the order of the integrals and combining the two phasor components of the second integral we obtain (in red) the shifted FTC of the signal $y(t)$. The resulting integral is the convolution integral of $X(f)$ with $Y(f)$.
\begin{eqnarray*}
\ldots&=& \int_{\eta=-\infty}^{\infty} X(\eta) \left (
\int_{t=-\infty}^{\infty} y(t) e^{j2 \pi \eta t} e^{-j2 \pi f t} \text{d} t
\right ) \text{d} \eta \\
&=& \int_{\eta=-\infty}^{\infty} X(\eta) \left ( \color{red}{
\int_{t=-\infty}^{\infty} y(t) e^{-j2 \pi (f- \eta) t} \text{d} t }
\right ) \text{d} \eta \\
&=& \int_{\eta=-\infty}^{\infty} X(\eta) \color{red}{Y(f-\eta)} \text{d} \eta
= X(f) \star Y(f)
\end{eqnarray*}
This leads to the following important “rule of thumb”:
Multiplication in one domain results in convolution in the other domain.
Example
As an example of the convolution property we will use the modulation property as discussed before. In this example signal $x(t)$ is a sinc function as depicted in the upper left side of the figure below.
Modulation in time domain results in frequency shifts of the spectrum.
Previously we have shown that the sinc function is the result of the IFTC of a frequency domain block function, as depicted on in the upper right side of the figure. Multiplying signal $x(t)$ with a sinusoidal signal with frequency $f_0$, which is a modulation operation, is according to the last "rule of thumb" equivalent to the convolution of the FTC $X(f)$ with the FTC of the sinusoidal. The FTC of the sinusoidal consists of two delta pulses, $\delta(f-f_0)$ and $\delta(f+f_0)$, as depicted in the middle part of the figure. Convolving $X(f)$ with a delta pulse results in a shift, which can be shown as follows:
\begin{eqnarray*}
\delta(f \pm f_0) \star X(f) &=& \int_{\eta=-\infty}^{\infty} \delta(\eta \pm f_0) X(f-\eta) \text{d} \eta = X(f \pm f_0)
\end{eqnarray*}
The result is shown in the lower part of the figure.
By using the IFTC definition of $x(t)$ and differentiating both sides with respect to $t$ we obtain:
$$
\frac{\text{d} }{\text{d} t} x(t) = \frac{\text{d} }{\text{d} t} \left (
\int_{-\infty}^{\infty} X(f) e^{j2 \pi f t} \text{d} f \right )
= \int_{-\infty}^{\infty} X(f) (j 2 \pi f) e^{j2 \pi f t} \text{d} f
= \mathcal{F}^{-1}\{j \omega X(f) \}
$$
Example
Find the FTC of the signal $x(t)= \frac{\text{d} }{\text{d} t} \delta(t)$.
Taking the FTC of both sides gives:
$$
X(f)= \mathcal{F}\{\frac{\text{d} }{\text{d} t} \delta(t)\} = j \omega \cdot \mathcal{F}\{ \delta(t) \} = j \omega \cdot 1 = j \omega
$$
In the first step of the proof we replace the complex conjugated version of $x(t)$ by the IFTC of the FTC of $x^\ast(t)$.
\begin{eqnarray*}
\int_{-\infty}^{\infty} |x(t)|^2 \text{d} t &=& \int_{-\infty}^{\infty} x(t) \color{red}{x^\ast(t)} \text{d} t
= \int_{-\infty}^{\infty} x(t) \mathcal{F}^{-1}\{ \color{red}{\mathcal{F} \{x^\ast(t)\}}\} \text{d} t
\end{eqnarray*}
Using the FTC property of complex conjugation and then applying the IFTC to this result gives:
\begin{eqnarray*}
\int_{-\infty}^{\infty} |x(t)|^2 \text{d} t
&=&\int_{-\infty}^{\infty} x(t) \mathcal{F}^{-1}\{\color{red}{ X^\ast(-f)}\} \text{d} t \\
&=& \int_{-\infty}^{\infty} x(t) \left (
\int_{-\infty}^{\infty} X^\ast(-f) e^{j2 \pi f t} \text{d} f \right ) \text{d} t
\end{eqnarray*}
Finally interchanging the variable $-f$ by $f$ and then interchanging the order of the integrals and using the FTC definition leads to the final result.
\begin{eqnarray*}
\int_{-\infty}^{\infty} |x(t)|^2 \text{d} t
&=& \int_{-\infty}^{\infty} x(t) \left (
\int_{-\infty}^{\infty} X^\ast(f) e^{-j2 \pi f t} \text{d} f \right ) \text{d} t \\
&=& \int_{-\infty}^{\infty} X^\ast(f) \left ( \color{red}{
\int_{-\infty}^{\infty} x(t) e^{-j2 \pi f t} \text{d} t } \right ) \text{d} f \\
&=& \int_{-\infty}^{\infty} |X(f)|^2 \text{d} f
\end{eqnarray*}
Concluding it follows that the theorem of Parseval states that the energy in time- and frequency-domain is the same. This implies that the FTC cannot introduce or lose energy, which leads to the following “rule of thumb”:
Energy in one domain is the same as energy in the other domain.
Example
In this example the goal is to calculate the energy of a time domain sinc function as depicted on the left side of the figure below.
When using the definition of a sinc function we can calculate the energy as follows:
\begin{eqnarray*}
E &=& \int_{t=-\infty}^{\infty} |{\color{red}x(t)}|^2 \text{d} t
=\int_{t=-\infty}^{\infty} |2 F_c \text{sinc}(2 \pi F_c t)|^2 \text{d} t \\
&=& \int_{t=-\infty}^{\infty} |2 F_c \frac{\sin(2 \pi F_c t)}{2 \pi F_c t}|^2 \text{d} t
\end{eqnarray*}
The evaluation of this integral is not trivial. However, when using the theorem of Parseval we can find the same result by evaluating the energy in frequency domain. As shown before the FTC of a time domain sinc function is a block function in frequency domain, as depicted on the right side of the figure above and the result is as follows:
\begin{eqnarray*}
E &=& \int_{t=-\infty}^{\infty} |{\color{red}x(t)}|^2 \text{d} t \\
&\overset{\text{Parseval}}{=} & \int_{f=-\infty}^{\infty} |{\color{blue}X(f)}|^2 \text{d} f = \int_{f=-F_c}^{F_c} |1|^2 \text{d} f = 2 F_c
\end{eqnarray*}
