Concept explainers
(a)
Find the Fourier transform of
(a)

Answer to Problem 35P
The Fourier transform of
Explanation of Solution
Given data:
Formula used:
Consider the general form of Fourier transform of
Consider the general form of inverse Fourier transform of
Consider scaling property.
Consider Time shift property.
Calculation:
Modify equation (1) as follows.
Substitute
From scaling and time shift property, equation (1) can be write as follows.
Substitute
Conclusion:
Thus, the Fourier transform of
(b)
Find the Fourier transform of
(b)

Answer to Problem 35P
The Fourier transform of
Explanation of Solution
Given data:
Formula used:
Consider Modulation property.
Calculation:
Modify equation (1) as follows.
Substitute
From modulation property, equation can be write as follows.
Conclusion:
Thus, the Fourier transform of
(c)
Find the Fourier transform of
(c)

Answer to Problem 35P
The Fourier transform of
Explanation of Solution
Given data:
Formula used:
Consider Time differentiation property.
Calculation:
Modify equation (1) as follows:
Substitute
From modulation property, equation can be write as follows:
Conclusion:
Thus, the Fourier transform of
(d)
Find the Fourier transform of
(d)

Answer to Problem 35P
The Fourier transform of
Explanation of Solution
Given data:
Formula used:
Consider Convolution in t property.
Calculation:
Modify equation (1) as follows.
Substitute
From Convolution in t property, equation can be write as follows.
Conclusion:
Thus, the Fourier transform of
(e)
Find the Fourier transform of
(e)

Answer to Problem 35P
The Fourier transform of
Explanation of Solution
Given data:
Formula used:
Consider Frequency differentiation property.
Calculation:
Modify equation (1) as follows.
Substitute
From Frequency differentiation property, equation can be write as follows.
Simplify the equation as follows.
Conclusion:
Thus, the Fourier transform of
Want to see more full solutions like this?
Chapter 18 Solutions
EE 98: Fundamentals of Electrical Circuits - With Connect Access
- A DPSK has the following data input: d(n) =101011010001 1. Find the output coded sequence and the carrier phase. 2. Recover the input data from the output coded sequence.arrow_forwardQ9 A single-phase transformer, 2500 / 250 V, 50 kVA, 50 Hz has the following parameters, the Primary and secondary resistances are 0.8 ohm and 0.012 ohm respectively, the primary and secondary reactance are 4 ohm and 0.04 ohm respectively and the transformer gives 96% maximum efficiency at 75% full-load. The magnetizing component of-load current is 1.2 A on 2500 V side. 1- Draw the equivalent circuit referred to primary (H.V side) and inserts all the values in it 2- Find out Ammeter, voltmeter and wattmeter readings on open-circuit and short-circuit test. If supply is given to 2500 V side in both cases. Ans. O.C. Test (Vo= 2500 V, lo=1.24 A, Wo=781.25 w) S.C. Test (Vsc =164.924 V, Isc =20 A, Wsc =800 w )arrow_forwardA modulating signal f(t) is bandlimited to 5.5 kHz is sampled at a rate of 15000 samples/sec. The samples are quantized into 1024 levels. Calculate transmission bandwidth if the following modulation types are used for signal transmission: 1-ASK, QAM 2-QPSK, 8-PSK 3-FSK, 8-FSK with Af = 20 kHzarrow_forward
- Q10 The full-load copper loss on the H.V. side of 100KVA, 11000/317 V, single-phase transformer is 0.62 kw and on the L.V. side is 0.48 kW. i) Calculate R1, R2 in ohms ii) Find X1,X2,if the percentage equivalent reactance is 4%, and reactance is divided in same proportion as resistance. Ans, 27.30, 0.175), 0.00482 . (7.5)arrow_forwardFind the binary sequence, for the following Differential Manchester code.arrow_forwardQ2- What are the parameters and loss that can be determined during open-circuit test of singlephase transformer. Draw the circuit diagram of open-circuit test and explain how can you calculate the Parameters and loss.arrow_forward
- Q6- the open circuit and short circuit tests on a 10 KVA, 125/250 v, 50 Hz single phase transformer gave the following results: O.C. Test: 125 V,0.6 A, 50 W ( on L.V.) S.C. Test: 20 V, 40 A, 177.78 W (on H.V. side) Calculate: i) Copper losses on half load ii) Full load efficiency at 0.8 leading p.f. iii) Half load efficiency at 0.8 leading p.f. iv) Regulation at full load at 0.9 leading p.f. Ans: 44.445 W, 97.23%, 97.69%, -1.8015%arrow_forwardQ3-A two-winding transformer has a primary winding with 208 turns and a secondary winding with 6 turns. The primary winding is connected to a 4160V system. What is the secondary voltage at no load? What is the current in the primary winding with a 50-amp load connected to the secondary winding? What is the apparent power flowing in the primary and secondary circuits? Ans. 120 V, 1.44 A, 6000 VAarrow_forwardQ12- A three phase transformer 3300/400 V,has D/Y connected and working on 50Hz. The line current on the primary side is 12A and secondary has a balanced load at 0.8 lagging p.f. Determine the i) Secondary phase voltage ii) Line current iii) Output power Ans. (230.95 V, 99.11 A, 54.94 kW)arrow_forward
- Q1- A single phase transformer consumes 2 A on no load at p.f. 0.208 lagging. The turns ratio is 2/1 (step down). If the loads on the secondary is 25 A at a p.f. 0.866 lagging. Find the primary current and power factor.arrow_forwardQ7- A 5 KVA, 500/250 V,50 Hz, single phase transformer gave the following reading: O.C. Test: 250 V,2 A, 50 W (H.V. side open) S.C. Test: 25 V10 A, 60 W (L.V. side shorted) Determine: i) The efficiency on full load, 0.8 lagging p.f. ii) The voltage regulation on full load, 0.8 leading p.f. iii) Draw the equivalent circuit referred to primary and insert all the values it.arrow_forwardQ4- A single phase transformer has 350 primary and secondary 1050 turns. The primary is connected to 400 V,50 Hz a.c. supply. If the net cross sectional area of core is 50 cm2, calculate i) The maximum value of the flux density in the core. ii) The induced e.m.f in the secondary winding. Ans: 1.029 T, 1200Varrow_forward
- Introductory Circuit Analysis (13th Edition)Electrical EngineeringISBN:9780133923605Author:Robert L. BoylestadPublisher:PEARSONDelmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage LearningProgrammable Logic ControllersElectrical EngineeringISBN:9780073373843Author:Frank D. PetruzellaPublisher:McGraw-Hill Education
- Fundamentals of Electric CircuitsElectrical EngineeringISBN:9780078028229Author:Charles K Alexander, Matthew SadikuPublisher:McGraw-Hill EducationElectric Circuits. (11th Edition)Electrical EngineeringISBN:9780134746968Author:James W. Nilsson, Susan RiedelPublisher:PEARSONEngineering ElectromagneticsElectrical EngineeringISBN:9780078028151Author:Hayt, William H. (william Hart), Jr, BUCK, John A.Publisher:Mcgraw-hill Education,





