Signals with Multiple Frequency Components

In this lab, the magnitude response of an electrical circuit is calculated  and this results is used 

to infer the effect of the circuit on some relatively complex input signals. In particular, the 

following input signal types  are applied to the circuit:

·       A signal composed of multiple sinusoidal waves of different frequencies

20[sin(1000πt) + sin(2000πt) + sin(20,000πt)]

·       A sinusoidal signal with a time-varying frequency (a sinusoidal sweep)

Figure 1: Calculation of Output Voltage

Figure 2: Calculation of Transfer Function

Figure 3: Multiple Sinusoidal Waves with Different Frequencies 

Figure 4: A Sinusoidal Sweep Wave 

Figure 5: Response of Circuit 

When frequency goes toward infinity, Xc goes toward zero. This means (Vout/Vin) will go 

to zero. When frequency goes toward zero, Xc goes toward infinity. This means (Vout/Vin) 

will go toward half. The response of circuit by H(ω) will determine.

Figure 6: A Schematic of the Circuit

Op Amp Relaxation Oscilator

        The first relaxation oscillator circuit, the astable multivibrator, was invented by Henri Abraham and Eugene Bloch using vacuum tubes during World War I. Balthasar van der Pol first distinguished relaxation oscillations from harmonic oscillations, originated the term "relaxation oscillator", and derived the first mathematical model of a relaxation oscillator, the influential Van der Pol oscillator model, in 1920. Van der Pol borrowed the term relaxation from mechanics; the discharge of the capacitor is analogous to the process of stress relaxation, the gradual disappearance of deformation and return to equilibrium in a inelastic medium. Relaxation oscillators can be divided into two classes.

 A) Saw tooth, sweep, or fly back oscillator.            B) Astable Multivibrator

Figure 1: A Relaxation Oscillator

Figure 2: A Relaxation Oscillator

Figure 3: Output Voltage and Capacitor Voltage

According to above graph frequency is 166.6 Hz.

percent frequency error = [(166.6 - 158)/166.6]*100% = 5.16%

Resistors have a little error and those effect on the other parameters.

Figure 4: A schematic of Relaxation Oscillator

Phasors Passive RL Circuit Response

          In this lab, the steady-state response of an electrical circuits is determined with a 

sinusoidal inputs.The input and output signals both have the same frequency, but the two

 signals can have different amplitudes and phase angles. The analysis of the circuit can be simplified 

by representing the sinusoidal signals as phasors. 

     Elements of series RL circuit are : L = 1mH and R = 47W.Input voltage frequencies:·        w = w_c /10 (low frequency input)
·        w = 10*w_c  (high frequency input)
·        w = w_c  (corner frequency input)

Figure 1: A Series RL Circuit

When frequency increasing the inductor voltage increasing and also phase difference between VL 

and VR, or between VL and I increase because VR is constant under frequency changes. Voltage 

across resistor in in phase with current circuit.

Figure 2: RL series at 1 KHz 

The phase difference is so small in above figure at 1 KHz.

Figure 3: RL Series at 7.1 KHz

The phase difference is so small in above figure at 1 KHz. According to the above grapg, frequency 

is equal with 7.1 KHz.  f = 1 / (140 μs) = 7143 Hz

Also ∆t = (2.5)*(5 μs) = 7.5 μs so ∆φ = (∆t / T) *(360°) , ∆φ = (7.5 / 140)*(360°) = 19.29°

 

Figure 4: A Schematic of  a Series RL Circuit

Class Activity (Week 12)

        Instantaneous, effective, and average power were discussed.

v(t) = Vm cos(ωt + θv) ,  i(t) = Im cos(ωt + θi)  or v(t) = Vm cos(ωt + θv - θi ) ,  i(t) = Im cos(ωt)

Instantaneous power is   P(t) = v(t)*i(t)  ,  P(t) = Pavg + Pavg cos(2ωt) - Pavg sin(2ωt) 

Average (Real) Power Pavg =  [(Vm*Im)/2] cos(θv - θi) = Vrms Irms cos(θv - θi) = S cos(θv - θi)

S is apparent power. Pf = Pavg / S = cos(θv - θi)

Reactive Power Q = [(Vm*Im)/2] sin(θv - θi)   

For resistive loads Q = 0 (Pf = 1) , For capacitive loads Q ˂ 0 (leading Pf) , 

For inductive loads Q  ˃ 0 (lagging Pf)

Figure 1: Effective or RMS Power

Vrms = Vm / √2   ,  Irms = Im / √2

Figure 2: Average Power

Figure 3: Apparent Power

S is apparent power. Pf = Pavg / S = cos(θv - θi)

Power Factor  Pf = cos(θv - θi)  ,  Reactive Power rf = sin(θv - θi) 

Lagging power factor shows that current lags voltage - hence an inductive load. 

Leading power factor shows that current leads voltage - hence a capacitive load.

Figure 4: Complex Power

Complex Power S (VA) = P (W) + jQ(VAR)  

P = (Ieff)^2 . R = (Veff)^2  / R   , Q = (Ieff)^2 . X = (Veff)^2  / X 

S = (Vrms).(Irms)* S = (V.I*)/2    ,    S = (Vrms * Irms) ˂(θv - θi) 

Figure 5: Power Triangle
(θv - θi) ˃ 0  Lagging power factor  ,  (θv - θi) ˃ 0  Lagging ˂ 0 Leading power factor

Figure 6: Impedance and Angel Phase

The power factor angle is equal to the angle of the load impedance, if V is the voltage across the load 

and I is current through the load.

Z = V˂θv  / I˂θi = (Vm / Im) ˂(θv - θi)  ,    Z = (Vrms / Irms) ˂(θv - θi)  

Figure 7: An Inductor Load (Lagging Power Factor)

Q = ((0.707*210)^2) / (2π*50*0.5) = 140.4 VAR

Figure 8: An Example

FREQUENCY DEPENDANCE AND TRANSFER FUNCTIONS

        The frequency response of a circuit may also be considered as the variation of the gain and phase with

frequency. The transfer function H(ω) (network function) is a useful analytical tool for finding

the frequency response of a circuit. In fact, the frequency response of a circuit is the plot of the circuit’s

transfer function H(ω) versus ω, with ω varying from ω = 0 to ω = ∞. 

H(ω) = Voltage gain = Vo(ω) / Vi(ω)    ,  H(ω) = Current gain = Io(ω) / Ii(ω) 

H(ω) = Transfer Impedance = Vo(ω) / Ii(ω)   , H(ω) = Transfer Admittance = Io(ω) / Vi(ω)  

 H(ω) = H(ω) φ. H(ω) = N(ω) / D(ω) . 

The roots of transfer function are determined when N(ω) = 0.  

 The poles of transfer function are determined when D(ω) = 0.

Figure 9: Zeros and Poles of a Series RLC

Figure 10: Zeros and Poles of a Parallel Circuit

Figure 11

Class Activity (Week 11)

   This week node analysis, mesh analysis, and superposition analysis are use for circuits that contain 

resistors, inductors, and capacitors.

What is Vc in the below circuit?

Figure 1: Node Analysis and an Dependent-Current Source

Figure 2

Also source transformation is practiced again. 

 Thevenin and Norton equivalent circuits are a source transformation.

Figure 3: Thevenin and Norton equivalent circuits 

In order to find the Zth in a circuit, independent current sources must removed (open circuit), and 

independent voltage sources must short circuit. Then an arbitrary fixed current source or an arbitrary

fixed voltage source must connected to the points a and b.

Figure 4

Instantaneous and Average Power is discussed. 

  v(t) = V _m cos(ωt + θ _v )  ,  i(t) = I _m cos(ωt + θ _i )

Where V_m and I_m are the maximum values of v(t) and i(t), also θ_v and θ_i are the phase angles of the

voltage and current, respectively. The instantaneous power absorbed by the circuit is as below

formula.

p(t) = v(t)*i(t) , P(t) =  V _m I _m cos(ωt + θ _v ) cos(ωt + θ _i ) 

Pavg = ½ Re[VI*] = ½ V _m I _m cos(θ _v − θ _i )

According to average power, a resistor absorbs only power, while a inductor or a capacitor does not.

                                  
                                        MAXIMUM AVERAGE POWER TRANSFER


If Vth and Zth are used insted of a circuit, Maximum average power will occur under below 

conditions: 

Zth = Rth + j Xth     ,     Z_L = R _L + jX _L

Z_L = R _L + jX _L = R _Th − jX _Th = Z*_Th

The Maximum Average Power transferred is:        Pmax = |Vth|^2 / (8*Rth)   

Measuring Gain and Phase Difference

           A series circuit of two resistors are connected to a sinusoidal voltage, and its phase difference

between input and output is measured by a oscilloscope.

V(t) = Vm sin(ωt + Φ)  , Vm = 2v , Φ = 0°
 

Figure 1: Schematic of Series Circuit of two Resistors

Figure 2: Series Resistor Circuit with 1 KHz Frequency

f = 1/T , f = 1/(1 ms)  , f = 1 KHz

Figure 3: Series Resistor Circuit with 5 KHz Frequency

f = 1/ (200 μs) , f = 5 KHz

Figure 4: Series Resistor Circuit with 10 KHz Frequency

f = 1/T , f = 1/(100μs)  ,  f = 10 KHz

There are no phase difference between input and output in a pure resistive circuit.

Rtotal = R1 + R2 , Rtotal = 45.9 + 97.7 = 143.6 Ω

According to fig 2 and scale, 50 mA = 40 mm → I(total) = (10 mm)*(50 mA/40 mm) = 12.5 mA


Vout = R2*I(total)  , Vout(Measure) = (97.7Ω )*(12.5 mA) = 1.22 v 

G = Vout/Vin , G = 1.22/2 , G = 0.61

G(Theory) = Vout/Vin = R2/(R1+R2) , G(Theory) = 97.7/(97.7+45.9) = 0.68 

Gain Percent Error = [(0.68-0.61)/0.68]*100% = 10.3%

Figure 5: Series RL Circuit with an AC Voltage

Figure 6: Graph of Voltage and Current in an Series RL Circuit with 1 KHz Input Voltage

According to the Figure 6 and scale, Vout = (24 mm)*(200 mv/ 20 mm) = 240 mv (Blue graph)

VR1 = (35 mm)*(1v/ 20 mm) = 1.75 v , I(total) = (16 mm)*(50 mA/ 20 mm) = 40 mA

Inductor voltage (blue graph) leads current (small red graph) by 90 degree , and two red graph are in

phase because those show current and voltage of R1.

|XL| = (2π*1000)*(1mH) = 2π Ω , G(Measure) = Vout/Vin , G = 0.24/2 = 0.12 

G(Theory) = XL/Z , Z = 45.9 + j(2π) , G(Theory) = 0.136 ˂ 82.2°


Gain Percent Error = [(0.136 - 0.12)/0.136]*100% = 11.8%

Figure 7: Graph of Voltage and Current in an Series RL Circuit with 5 KHz Input Voltage

According to the Figure 7 and scale, Vout = (40 mm)*(500 mv/ 20 mm) = 1 v (Blue graph)

VR1 = (32 mm)*(1v/ 20 mm) = 1.6 v , I(total) = (14 mm)*(50 mA/ 20 mm) = 35 mA

Inductor voltage (blue graph) leads current (small red graph) by 90 degree , and two red graph are in

phase because those show current and voltage of R1.

 |XL| = (2π*5000)*(1mH) = 10π Ω  , G(Measure) = Vout/Vin , G = 1/2 = 0.5 

G(Theory) = XL / Z , Z = 45.9 +j(10π) , G(Theory) = 0.565 ˂ 55.6°

When frequency increases, XL, Vout, and G increases and vice versa.

Figure 8: A Practical Series RC Circuit

Figure 9: A Series RC Circuit

Figure 10: A Series RC Circuit with 1 KHz Frequency

According to the Figure 10 and scale, Vout = (29mm)*(1v/ 30 mm) = 0.97v (Blue graph)

 |XC| = 1/(2π*1000)*(0.41μF) = 388.4 Ω    ,   G(Measure) = Vout/Vin  , G = 0.97/2 = 0.485

 Z = 46.1 - j(388.4) , G(Theory) = XC / Z , G(Theory) = 0.993 ˂ -6.8°

Gain Percent Error = [(0.993 - 0.485)/0.993]*100% = 51.2%

Figure 11: A Series RC Circuit with 5 KHz Frequency

According to the Figure 11 and scale, Vout = (35 mm)*(1v/ 30 mm) = 1.17v (Blue graph)

 |XC| = 1/(2π*5000)*(0.41μF) = 77.68 Ω    ,   G(Measure) = Vout/Vin  , G = 1.17/2 = 0.585

G(Theory) = XC / Z , Z = 46.1 - j(77.68) , G(Theory) = 0.859 ˂ -30.7°

Gain Percent Error = [(0.859 - 0.585)/0.859]*100% = 31.9%

Figure 12: A Series RC Circuit with 10 KHz Frequency

According to the Figure 12 and scale, Vout = (45 mm)*(1v/ 40 mm) = 1.13 v (Blue graph)

 |XC| = 1/(2π*10000)*(0.41μF) = 38.9 Ω    ,   G(Measure) = Vout/Vin  , G = 1.13/2 = 0.565

G(Theory) = XC / Z , Z = 46.1 - j(38.9) , G(Theory) = 0.645 ˂ -49.9°

Gain Percent Error = [(0.645 - 0.565)/0.645]*100% = 12.4%

When frequency increases, XC, Vout, and G decreases and vice versa.