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LAB REPORT

EE 361 – Applied Electromagnetics

Lab Number 1 Simple Frequency and Time Analysis

Names:

Hieu Nguyen Iman Daneshnia

Section Number AC & AB

I.

Procedure 1

We used the Pspice to build the schematic for the given circuit as shown in the figure 1 below:

Figure 1: Schematic of first circuit

The values of the component are as shown in the fiqure 1, were given in the lab manual. Then, we chose the frequency in a range of 100MEG to 10G. Then, we monitor our output results and behavior of the circut due to Transmission Line, by plotting parameters such as current or voltage. The following graphs shows our output: Note: In all the following graphs, I(RS) corresponds to source current, Vsource corresponds to source voltage, I(RL) corresponds to load current, and Vload corresponds to load voltage. Source Current and Voltage Graphs: The Figure 2 below showes the plot of the Source current. As frequency increase, the current amplitude varies less, and becoming almost DC with approximate value of 10 mA.

Figure 2: Source current

The Figure 3 below showes the plot of the source voltage. As frequency increase, the voltage amplitude varies less, and becoming almost DC with approximate value of 500 mV.

Figure 3: Voltage source The picture make sense, because we know that the source will have the wave form of cosine function, and as the frequency increases, the pk-to-pk amplitude in the plot varies less, became close to DC. Load Current and Voltage Graphs: The Figure 4 below showes the plot of the load current. As frequency increase, the current amplitude decay up,and reaches an almost constant value of 10 mA. Since the wave form of the current is too small compare to voltage amplitude, we decided to make seperate plot for cuurent and voltage.

Figure 4: Load current The Figure 5 below showes the plot of the load voltage. As frequency increase, the voltage amplitude decay down,and reaches an almost constant value of 500 mV.

Figure 5: load voltage

M()-Magnitude of Source Current and Voltage:

Figure 6: Magnitude current source

Figure 7: Magnitude voltage source The Figure 8 and 9, above, show the Magnitude of current source and voltage source. As before, by increasing the frequencies, the amplitude decays out, still sinusoidal, but decreasing pl-to-pk value.

P() - Phase of Source Current and Voltage: As you can see in the Figure 8 and 9, below; the current source and voltage source are out of phase.

Figure 8: Phase current source

Figure 9:Phase voltage source

M()-Magnitude of Load Current and Voltage: The Figure 10 shows the Magnitude of the load current. As frequency increases, the current decay up exponentioaly and reaches a constant value of 10 mA.

Figure10: Magnitude load current The Figure 11 shows the Magnitude of the load voltage. As frequency increases, the current decay down exponentioaly and reaches a constant value of 500 mV.

Figure 11: Magnitude load voltage

P() - Phase of Load Current and Voltage: The Figure 12 and 13 show the phasor plot of the current source and voltage source. As seen in the plots, both current and voltage are in phase with each other at the load side.

Figure 12: Phase load current

Figure 13: Phase load voltage

BODE-type of voltage source and voltage load: Then, we used the function 20*LOG10( ) to make plotted the BODE-type plot of voltage source and voltage load, which is shown in Figure 14 below.

Figure 14: BODE_plot Voltage source

Figure 15: BODE_plot Voltage Load

Then, we repeat the simulation for frequencies 100kHz to 100MHz. The figures 16 and 17 below, show our new BODE-type plots of voltage source and voltage load for the new simulation. As seen in the plots, from 100KHz to 300Khz, our voltage source and load voltage has value of almost zero.

Figure 16: Changed Frequency bode-plot Voltage source

Figure 17: Changed Frequency bode-plot Voltage Load

In this passive network, how can the voltage at the load be higher than the voltage at the source?

We can see from the picture that first the voltage load and voltage source are equals, but with increasing the frequency, the voltage load values gets higher than voltage source. The reason for this increase in the load voltage is due to capacitane of the Tranmission Line of the circut. As we know from the book and discussion in class, the tranmission line have impedance of inductance and capacitance. When the frequency increases, the capacitance decreases, because the capacitance of the voltage source is decreasing along with it. As a result, the voltage load will be slightly greater than voltage source.

II.

Procedure 2

In this part, we replaced the Vac to VSIN, with amplitude of 1 volt, and DC offset of zero volt. Our new circutschematic can be seen in Figure 18 below:

Figure 18: Schematic with new source

We ran a Time Domain Response (Transient) simulation. We use the transient analysis to obtain plots of the transient voltage waveforms VSOURCE and VLOAD for 5 periods of the wave. The resulting output can be seen in the folllowing Figure 19 and 20 below.

Figure 19: Transient voltage source

Figure 20: Transient voltage load

What is the phase delay (in degrees) at 0.5 GHz due to the transmission line? The phase delay at 0.5 GHz due to the transmission line can be calcultaed using the following equation: t Delay Delay 360 TPeriod

tDelay=1 nS TPeriod=2 nS Ө Delay=(1/2)(360)=180 degree

Does this make sense, given the transmission line parameters? Yes. It does makes sense. It is because, we set the frequency of the line to be 0.5GHz, therefore, the period for that will be: T = 1/(0.5*109) = 2ns We also can verify our period by looking at the plot, which has the same time period as we calculated.

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EE 361 – Applied Electromagnetics

Lab Number 1 Simple Frequency and Time Analysis

Names:

Hieu Nguyen Iman Daneshnia

Section Number AC & AB

I.

Procedure 1

We used the Pspice to build the schematic for the given circuit as shown in the figure 1 below:

Figure 1: Schematic of first circuit

The values of the component are as shown in the fiqure 1, were given in the lab manual. Then, we chose the frequency in a range of 100MEG to 10G. Then, we monitor our output results and behavior of the circut due to Transmission Line, by plotting parameters such as current or voltage. The following graphs shows our output: Note: In all the following graphs, I(RS) corresponds to source current, Vsource corresponds to source voltage, I(RL) corresponds to load current, and Vload corresponds to load voltage. Source Current and Voltage Graphs: The Figure 2 below showes the plot of the Source current. As frequency increase, the current amplitude varies less, and becoming almost DC with approximate value of 10 mA.

Figure 2: Source current

The Figure 3 below showes the plot of the source voltage. As frequency increase, the voltage amplitude varies less, and becoming almost DC with approximate value of 500 mV.

Figure 3: Voltage source The picture make sense, because we know that the source will have the wave form of cosine function, and as the frequency increases, the pk-to-pk amplitude in the plot varies less, became close to DC. Load Current and Voltage Graphs: The Figure 4 below showes the plot of the load current. As frequency increase, the current amplitude decay up,and reaches an almost constant value of 10 mA. Since the wave form of the current is too small compare to voltage amplitude, we decided to make seperate plot for cuurent and voltage.

Figure 4: Load current The Figure 5 below showes the plot of the load voltage. As frequency increase, the voltage amplitude decay down,and reaches an almost constant value of 500 mV.

Figure 5: load voltage

M()-Magnitude of Source Current and Voltage:

Figure 6: Magnitude current source

Figure 7: Magnitude voltage source The Figure 8 and 9, above, show the Magnitude of current source and voltage source. As before, by increasing the frequencies, the amplitude decays out, still sinusoidal, but decreasing pl-to-pk value.

P() - Phase of Source Current and Voltage: As you can see in the Figure 8 and 9, below; the current source and voltage source are out of phase.

Figure 8: Phase current source

Figure 9:Phase voltage source

M()-Magnitude of Load Current and Voltage: The Figure 10 shows the Magnitude of the load current. As frequency increases, the current decay up exponentioaly and reaches a constant value of 10 mA.

Figure10: Magnitude load current The Figure 11 shows the Magnitude of the load voltage. As frequency increases, the current decay down exponentioaly and reaches a constant value of 500 mV.

Figure 11: Magnitude load voltage

P() - Phase of Load Current and Voltage: The Figure 12 and 13 show the phasor plot of the current source and voltage source. As seen in the plots, both current and voltage are in phase with each other at the load side.

Figure 12: Phase load current

Figure 13: Phase load voltage

BODE-type of voltage source and voltage load: Then, we used the function 20*LOG10( ) to make plotted the BODE-type plot of voltage source and voltage load, which is shown in Figure 14 below.

Figure 14: BODE_plot Voltage source

Figure 15: BODE_plot Voltage Load

Then, we repeat the simulation for frequencies 100kHz to 100MHz. The figures 16 and 17 below, show our new BODE-type plots of voltage source and voltage load for the new simulation. As seen in the plots, from 100KHz to 300Khz, our voltage source and load voltage has value of almost zero.

Figure 16: Changed Frequency bode-plot Voltage source

Figure 17: Changed Frequency bode-plot Voltage Load

In this passive network, how can the voltage at the load be higher than the voltage at the source?

We can see from the picture that first the voltage load and voltage source are equals, but with increasing the frequency, the voltage load values gets higher than voltage source. The reason for this increase in the load voltage is due to capacitane of the Tranmission Line of the circut. As we know from the book and discussion in class, the tranmission line have impedance of inductance and capacitance. When the frequency increases, the capacitance decreases, because the capacitance of the voltage source is decreasing along with it. As a result, the voltage load will be slightly greater than voltage source.

II.

Procedure 2

In this part, we replaced the Vac to VSIN, with amplitude of 1 volt, and DC offset of zero volt. Our new circutschematic can be seen in Figure 18 below:

Figure 18: Schematic with new source

We ran a Time Domain Response (Transient) simulation. We use the transient analysis to obtain plots of the transient voltage waveforms VSOURCE and VLOAD for 5 periods of the wave. The resulting output can be seen in the folllowing Figure 19 and 20 below.

Figure 19: Transient voltage source

Figure 20: Transient voltage load

What is the phase delay (in degrees) at 0.5 GHz due to the transmission line? The phase delay at 0.5 GHz due to the transmission line can be calcultaed using the following equation: t Delay Delay 360 TPeriod

tDelay=1 nS TPeriod=2 nS Ө Delay=(1/2)(360)=180 degree

Does this make sense, given the transmission line parameters? Yes. It does makes sense. It is because, we set the frequency of the line to be 0.5GHz, therefore, the period for that will be: T = 1/(0.5*109) = 2ns We also can verify our period by looking at the plot, which has the same time period as we calculated.

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