Understanding Sinusoidal Functions and Phasor Representation
Added on 2022-12-23
13 Pages1732 Words20 Views
Question 1
a) Sin(2t+2)-1
F(t)=X*sin(Y(t)+Z) - A
A function in the form shown above is explained as follows
Multiplier X gives the amplitude of the sine wave thus from our question the amplitude is
1 unit
To find the period, ω (omega) of the sine waveform, Y is used as shown in the formula
below
ω=period of regular sine wave/|Y|, but period of regular sine wave is 2*∏
= 2*∏/2
= ∏
Z/Y from the equation gives the phase shift of the function, from our question the sine
wave is shifted to the left (due to positive sign that comes before Z) by one unit. A on the
other hand indicates the vertical shift of the waveform, negative sign before A represent
downward shift and a positive sign upward shift, thus the function in question was shifted
downwards by 1 unit
To find the frequency, F of a function
F=1/period, period was found to be ∏
=1/∏ hertz
RMS, instantaneous, peak to peak and average value is cleared demonstrated by Figure 1:
Figure 1: properties of a waveform
Instantaneous value of the function = Vmax*sin (ω) t, omega was found to be ∏,
thus
a) Sin(2t+2)-1
F(t)=X*sin(Y(t)+Z) - A
A function in the form shown above is explained as follows
Multiplier X gives the amplitude of the sine wave thus from our question the amplitude is
1 unit
To find the period, ω (omega) of the sine waveform, Y is used as shown in the formula
below
ω=period of regular sine wave/|Y|, but period of regular sine wave is 2*∏
= 2*∏/2
= ∏
Z/Y from the equation gives the phase shift of the function, from our question the sine
wave is shifted to the left (due to positive sign that comes before Z) by one unit. A on the
other hand indicates the vertical shift of the waveform, negative sign before A represent
downward shift and a positive sign upward shift, thus the function in question was shifted
downwards by 1 unit
To find the frequency, F of a function
F=1/period, period was found to be ∏
=1/∏ hertz
RMS, instantaneous, peak to peak and average value is cleared demonstrated by Figure 1:
Figure 1: properties of a waveform
Instantaneous value of the function = Vmax*sin (ω) t, omega was found to be ∏,
thus
=1*sin (∏) t, where Vmax is the amplitude
RMS value of the function= peak voltage*1/ , peak value is 1 from the function
= 1*0.707
= 0.707 units
Peak to peak value is 2 units, one unit above reference axis and one unit below
Average value of the function = peak value* 0.637
= 1*0.637
= 0.637 units
b) Function 2 sin 5 t
F(t)= X sin Y (t)
X= 2, it represents the amplitude of the waveform
Y= 5, used to compute period of the waveform
Period, ω of the waveform is given by
ω=regular period of a sine function/Y
=2*∏/5= 2/5*∏
Frequency of the function is calculated from the period by the relation below
F= 1/period = 1 / (2∏/5) = 5/2∏
Instantaneous value of the sine function is
Vi= Vmax*sin ω t, from above computation Vmax is equal to the amplitude and
ω=2/5∏
Vi= 2*sin (2∏/5) t
Peak to peak value is 4 units, 2 units above reference axis and 2 units below
RMS value is given by peak value multiplied by 1/
Peak value of the function from figure 1 is given by 2 units
RMS value=peak value * 0.707
= 2*0.707
=1.414 units
Average value is computed by multiplying peak value by 0.637
Average value= 2*0.637
= 1.274 units
RMS value of the function= peak voltage*1/ , peak value is 1 from the function
= 1*0.707
= 0.707 units
Peak to peak value is 2 units, one unit above reference axis and one unit below
Average value of the function = peak value* 0.637
= 1*0.637
= 0.637 units
b) Function 2 sin 5 t
F(t)= X sin Y (t)
X= 2, it represents the amplitude of the waveform
Y= 5, used to compute period of the waveform
Period, ω of the waveform is given by
ω=regular period of a sine function/Y
=2*∏/5= 2/5*∏
Frequency of the function is calculated from the period by the relation below
F= 1/period = 1 / (2∏/5) = 5/2∏
Instantaneous value of the sine function is
Vi= Vmax*sin ω t, from above computation Vmax is equal to the amplitude and
ω=2/5∏
Vi= 2*sin (2∏/5) t
Peak to peak value is 4 units, 2 units above reference axis and 2 units below
RMS value is given by peak value multiplied by 1/
Peak value of the function from figure 1 is given by 2 units
RMS value=peak value * 0.707
= 2*0.707
=1.414 units
Average value is computed by multiplying peak value by 0.637
Average value= 2*0.637
= 1.274 units
Question 2
Amplitude is the measure of maximum displacement of a wave from its point of equilibrium.
Period is the time taken to complete one cycle.
Frequency is the measure of cycles of a wave in a second.
Peak-to-peak is the measure of the displacement from top crest to the bottom crest. It is also
equivalent to twice the amplitude.
Instantaneous value of an alternating waveform is the value of the wave at a particular instant.
The average value of a sinusoidal quantity is the average of half a cycle and is calculated by
multiplying peak value by a factor of 0.637.
The RMS value of a wave is the effective value of an alternating waveform and is equivalent to
peak value multiplied by 0.707
Question 3
Resistor value is 47 ohms and a 50 V source supply
Instantaneous value of the voltage source, e is given by the equation
e=Ep*sin ω (t)
Ep is the peak value of the voltage in volts,
ω is the angular velocity in radians
Ep= E rms* , E rms is the RMS value of the voltage and is provided in the question
= 50 *
=70.71 volts
Instantaneous value of the voltage source
e=70.71*sin ω t
voltage drop across the resistor is equal the voltage of source, thus a voltage drop across the
resistor is given by
Amplitude is the measure of maximum displacement of a wave from its point of equilibrium.
Period is the time taken to complete one cycle.
Frequency is the measure of cycles of a wave in a second.
Peak-to-peak is the measure of the displacement from top crest to the bottom crest. It is also
equivalent to twice the amplitude.
Instantaneous value of an alternating waveform is the value of the wave at a particular instant.
The average value of a sinusoidal quantity is the average of half a cycle and is calculated by
multiplying peak value by a factor of 0.637.
The RMS value of a wave is the effective value of an alternating waveform and is equivalent to
peak value multiplied by 0.707
Question 3
Resistor value is 47 ohms and a 50 V source supply
Instantaneous value of the voltage source, e is given by the equation
e=Ep*sin ω (t)
Ep is the peak value of the voltage in volts,
ω is the angular velocity in radians
Ep= E rms* , E rms is the RMS value of the voltage and is provided in the question
= 50 *
=70.71 volts
Instantaneous value of the voltage source
e=70.71*sin ω t
voltage drop across the resistor is equal the voltage of source, thus a voltage drop across the
resistor is given by
e=70.71*sin ω t
amplitude of voltage drop is given by 70.71 volts from the equation above
period of the waveform is ω= 2*∏*f = 2* *60=377
frequency is equal to that of the source and is assumed to be 60 Hz
peak to peak value of the voltage drop is given by 2*70.71=141.42 volts
instantaneous value of the voltage drop is given by
Vi=Vmax*sin ω t
= 70.71*sin (377) t
RMS value of the voltage drop is given by peak value multiplied by 1/
RMS value= 70.71*0.707
= 50 volts
Average value of the voltage drop is given by
Average value=peak voltage*0.637
= 45.04 volts
Question 4
Phasor representation of two sinusoidal voltages
amplitude of voltage drop is given by 70.71 volts from the equation above
period of the waveform is ω= 2*∏*f = 2* *60=377
frequency is equal to that of the source and is assumed to be 60 Hz
peak to peak value of the voltage drop is given by 2*70.71=141.42 volts
instantaneous value of the voltage drop is given by
Vi=Vmax*sin ω t
= 70.71*sin (377) t
RMS value of the voltage drop is given by peak value multiplied by 1/
RMS value= 70.71*0.707
= 50 volts
Average value of the voltage drop is given by
Average value=peak voltage*0.637
= 45.04 volts
Question 4
Phasor representation of two sinusoidal voltages
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