Exercise 3.4Z: Trapezoid, Rectangle and Triangle

Trapezoidal pulse and its limiting cases "Rectangle" and "Triangle"

Three different pulse shapes are considered. The pulse ${x(t)}$ is trapezoidal. For $| t | < t_1 = 4 \,\text{ms}$ the time course is constant equal to ${A} = 1\, \text{V}$. Afterwards, ${x(t)}$ drops linearly to the value zero until the time $t_2 = 6\, \text{ms}$.

The spectral function of the trapezoidal pulse is

$$X( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits}( {{\rm \pi} \cdot \Delta t \cdot f} ) \cdot \hspace{0.1cm}{\mathop{\rm si}\nolimits}( {{\rm \pi}\cdot \Delta t \cdot r_t \cdot f} ).$$

with the two derived system quantities, namely

the equivalent bandwidth,

$$\Delta t = t_1 + t_2,$$

and the so-called roll-off factor (in the time domain):

$$r_t = \frac{t_2 - t_1 }{t_2 + t_1 }.$$

Furthermore, the rectangular pulse ${r(t)}$ and the triangular pulse ${d(t)}$ are also shown in the graph, both of which can be interpreted as limiting cases of the trapezoidal pulse ${x(t)}$.

Hints:

This exercise belongs to the chapter Fourier Transform Theorems.
You can check your results using the two interactive applets

Pulses and Spectra,

Frequency & Impulse Responses.

Questions

$\Delta t \ = \ $

$\text{ms}$

$r_t\hspace{0.3cm} = \ $

	The spectral value at frequency $f = 0$ is equal to $20 \,\text{mV/Hz}$.
	For the phase function the values $0$ and $\pi$ $(180^{\circ})$ are possible.
	${X(f)}$ only has zeros at all multiples of $100 \,\text{Hz}$.

	The spectral value at frequency $f = 0$ is equal to ${X(f = 0)}$.
	The values $0$ and $\pi$ $(180^{\circ})$ are possible for the phase function.
	${R(f)}$ only has zeros at all multiples of $100 \,\text{Hz}$.

	The spectral value at frequency $f = 0$ is equal to ${X(f = 0)}$.
	The values $0$ and $\pi$ $(180^{\circ})$ are possible for the phase function.
	${D(f)}$ only has zeros at all multiples of $100 \,\text{Hz}$.

Solution

(1) The equivalent pulse duration is $\Delta t = t_1 + t_2 \;\underline{= 10 \,\text{ms}}$ and the rolloff factor is $r_t = 2/10 \;\underline{= 0.2}$.

(2) Proposed solutions 2 and 3 are correct:

The spectral value at $f = 0$ is $A \cdot \Delta t = 10 \,\text{mV/Hz}$.
Since ${X(f)}$ is real and can assume both positive and negative values, only the two phase values $0$ und $\pi$ are possible.
Zeros exist due to the first si–function at all multiples of $1/\Delta t = 100\, \text{Hz}$.
The second si–function leads to zero crossings at intervals of $1/(r_t \cdot \Delta t) = 500 \,\text{Hz}$. These coincide exactly with the zeros of the first si–function.

(3) All proposed solutions are correct:

With the equivalent pulse duration $\Delta t = 10 \,\text{ms}$ and the rolloff factor $r_t = 0$ one obtains: $R( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits} ( {{\rm{\pi }} \cdot \Delta t \cdot f} ).$
It follows that $R( f = 0) = A \cdot \Delta t = X( f = 0).$

(4) Proposed solutions 2 and 3 are correct:

For the triangular pulse, the rolloff factor is $r_t = 1$.
The equivalent pulse duration is $\Delta t = 10 \,\text{ms}$. It follows that $D( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits} ^2 ( {{\rm{\pi }} \cdot \Delta t \cdot f} )$ and $D( f = 0) = A \cdot \Delta t = X( f = 0)$.
Since ${D(f)}$ cannot become negative, the phase $[{\rm arc} \; {D(f)}]$ is always zero. The phase value $\pi$ $(180°)$ is therefore not possible with the triangular pulse.