Function of filter circuit - Coursework Example

Function of filter circuitA filter circuit server the purpose of producing restrictions on the bandwidth of frequencies for an alternating input signal and generate output with a narrower frequency- bandwidth. It also eliminates the alternating current ripple remaining in the output of the rectification of a diode-based alternating current circuit, leading to higher quality signal.(b) Gain on the circuitGain across two components is derived asGc = |Hout(jw)|Gc = |(Vout(jw) / Vin(jw) )|Gc = 1 / √ (1 + (wRC) 2)GR = |HR (jw)| GR = |VR (jw) / Vin (jw)|GR = wRC / √ (1 + (wRC)2)Question 2The function of this diagramThis diagram represents the relationship between the capacitor and a resistor in the circuit, if they are arranged in series.

The circuit assists in measuring the voltage across the capacitor, using Kirchhoff's law of the current. In this, the current that charges the capacitor has to be equal to the current passing through the resistor.Circuit in SeriesQuestion 3A circuit that performs the opposite finction is drawn belowCircuit in ParallelExplanationThis circuit is a parallel RC circuit. It is less in voltage than the series circuit because its output voltage is the same as the input voltage. This circuit therefore, does not operate as a filter to the input signal except when supplied by the current source.

Question 4Characteristics of ideal Operational AmplifierInfinite bandwidthThe ideal op-amp amplifies all signals from Direct Current to the greatest Alternating Current frequenciesReal op-amps has limited bandwidthInfinite open-loop voltage gainOpen-Loop Gain is infinite in ideal Op-amp amplifier. In real Op-Amp, the open loop gain is finite, ranging between 20000 and 200000.Zero noise contributionIdeal op-amp generates zero noise voltage from the internal parts.Real op-amp has several sources of noise, including semiconductor noise and resistive noise.

Zero output impedanceIdeal op-amp operates as a perfect source of internal voltage without any internal resistance. Real op-amps have output-impedance ranging between 100 - 20Ω.Question 5: Inverting Op-AmpDeriving the Gain (GAv)G (Av) = V out / V inG (Av) = Rf / RinVout = Vin * (- Rf / Rin)Question 6Output = 7.46KWEfficiency = 85%Actual Output = (85 / 100) * 7.46 = 6.341KWCurrent = V / R = 440 / 200 = 2.2 AmpsBack EMF = 6.342 * 8 / 2.2 = 23.06182Question 7: Torque Speed Graph for dc compound motorQuestion 8:Armature Resistance = 0.

5 ohmsVoltage = 120VCurrent = Voltage / Resistance Current = 120 / 0.5 = 240 Amps Power = IVPower = 240 * 120 = 28800 WattCurrent = Voltage / (240 - 40)Current = 120 / 200 = 0.6INew Power = 120 * 0.5 = 60 WattLost Power = 28800 – 60 = 28740WattQuestion 9: With the 6 bits, the number of steps = 2^12 – 1 = 4096 - 1 = 4095 steps of size 10 The full scale output = 4095 * 10 mV = 40.950VPercentage Resolution = (10mV / 40.95 V) * 10024.42002 % Question 10Full scale output = step size * Number of steps = 2.

55(2^8 - 1) = 650.25V11111111 converted to decimal number = 255Output Voltage for an input of 11111111 = 255 * 2.55 = 650.25VAnalog Input = 2.55V Digital output = 650.25VThe difference = 650.25 – 2.55 = 647.7V