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The value of the current, i ( t ) shown in the given circuit with the assumptions i ( 0 ) = 0 , R = 10 Ω and L = 0.5 H and using the Laplace transforms. The value of the current, i ( t ) is i ( t ) = ( − 5 + 100 t + 5 e − 20 t 0 < t < 2 e − 20 t ( 5 + e 40 ) t > 2 ) _ . Definition used: Laplace transform of derivatives: “The first and second derivatives of f ( t ) satisfy ℒ ( f ′ ) = s ℒ ( f ) − f ( 0 ) ⋯ ( 1 ) ℒ ( f ″ ) = s 2 ℒ ( f ) − s f ( 0 ) − f ′ ( 0 ) ⋯ ( 2 ) Formula (1) holds if f ( t ) is continuous for all t ≥ 0 and satisfies the growth restriction (2) and f ′ ( t ) is piecewise continuous on every finite interval on the semi axis t ≥ 0 . Similarly (2) holds if f ( t ) and f ′ ( t ) are continuous for all t ≥ 0 and satisfy the growth restriction and f ″ ( t ) is piecewise continuous on every finite interval on the semi axis t ≥ 0 .” Unit step function or Heaviside Function: The unit step function or the Heaviside function denoted by u ( t − a ) is defined as follows. u ( t − a ) = { 0 , if t < a 1 , if t > a } Second shifting theorem If f ( t ) has the transform F ( s ) , then the shifted function f ˜ ( t ) = f ( t − a ) u ( t − a ) = { 0 if t < a f ( t − a ) if t > a } has the transform e − a s F ( s ) . That is, if ℒ ( f ( t ) ) = F ( s ) then, ℒ ( f ( t − a ) u ( t − a ) ) = e − a s F ( s ) ⋯ ( 1 ) Calculation: Consider the given conditions as follows. i ( 0 ) = 0 , R = 10 Ω and L = 0.5 H Also, the value of voltage is given as shown below. v = { 200 t if 0 < t < 2 0 if t > 2 } Use the above definition of unit step function and rewrite the given voltage function in terms of unit step function as shown below. v ( t ) = 200 t ( 1 − u ( t − 2 ) ) = 200 t − 200 t u ( t − 2 ) It is known that the differential equation of a voltage of an RL circuit is L d I d t + R I = V ( t ) . Substitute the values R = 10 Ω and L = 0.5 H in the equation L d I d t + R I = V ( t ) as shown below. L d I d t + R I = V ( t ) 0.5 d I d t + 10 I = V ( t ) Take Laplace transform on the both sides of the above equation; apply definition of Laplace transform of derivatives as shown below. ℒ ( 0.5 d I d t + 10 I ) = ℒ ( V ( t ) ) 0.5 s ℒ ( I ) + 10 ℒ ( I ) = ℒ ( 200 t − 200 t u ( t − 2 ) ) ( 0.5 s + 10 ) ℒ ( I ) = 200 s 2 − 200 e − 2 s ( 1 s 2 + 2 s ) ( Multiply both sides by 2 ) ( s + 20 ) ℒ ( I ) = 2000 s 2 − 2000 ⋅ e − 2 s ( 1 + 2 s s 2 ) On further simplifications of the above equation the following is obtained. L ( I ) = 2000 s 2 ( s + 20 ) − 2000 ⋅ e − 2 s ( 1 + 2 s ) s 2 = 2000 s 2 ( s + 20 ) − 2000 e − 2 s s 2 ( s + 20 ) − 2000 e − 2 s ⋅ 2 s ( s + 20 ) = 2000 ( 1 − e − 2 s s 2 ( s + 20 ) − 2 e − 2 s s ( s + 20 ) ) = 2000 ( 1 − e − 2 s − 2 s e − 2 s s 2 ( s + 20 ) ) Now, apply partial fractions on the right hand side of the above equation as follows. 1 ( s 2 ) ( s + 20 ) = A s + B s 2 + C s + 20 1 ( s 2 ) ( s + 20 ) = A s ( s + 20 ) + B ( s + 20 ) + C s 2 ( s 2 ) ( s + 20 ) 1 = A s ( s + 20 ) + B ( s + 20 ) + C s 2 1 = A s 2 + 20 A s + B s + 20 B + C s 2 Equate the coefficient of s 2 on both sides of the above equation as follows. 0 = A + C A = − C Equate constant terms on both sides as follows. 1 = 20 B ( Divide both sides by 20 ) 1 20 = 20 B 20 1 20 = B Equate the coefficient of s on both sides of the above equation as follows. 0 = 20 A + B 0 = 20 A + 1 20 ( subtract 1 20 on both sides ) 0 − 1 20 = 20 A + 1 20 − 1 20 − 1 20 = 20 A ( Divide both sides by 20 ) − 1 400 = A Therefore, the value of C is 1 400 . Substitute the values of A , B and C in the equation 1 ( s 2 ) ( s + 20 ) = A s + B s 2 + C s + 20 as follows. 1 ( s 2 ) ( s + 20 ) = A s + B s 2 + C s + 20 = − 1 400 s + 1 20 s 2 + 1 400 ( s + 20 ) Thus, ℒ ( i ) = 2000 ( 1 − e − 2 s − 2 s e − 2 s ) ( − 1 400 s + 1 20 s 2 + 1 400 ( s + 20 ) ) . Rewrite the above as follows. ℒ ( i ) = 2000 ( 1 − e − 2 s − 2 s e − 2 s ) ( − 1 400 s + 1 20 s 2 + 1 400 ( s + 20 ) ) = ( 1 − e − 2 s − 2 s e − 2 s ) ( − 5 s + 100 s 2 + 5 s + 20 ) = ( − 5 s + 100 s 2 + 5 s + 20 ) − e − 2 s ( − 5 s + 100 s 2 + 5 s + 20 ) − 2 s e − 2 s ( − 5 s + 100 s 2 + 5 s + 20 ) Now, take inverse Laplace transform on both sides as follows. i ( t ) = ( ℒ − 1 ( − 5 s + 100 s 2 + 5 s + 20 ) − ℒ − 1 ( e − 2 s ( − 5 s + 100 s 2 + 5 s + 20 ) ) − 2 ℒ − 1 ( s e − 2 s ( − 5 s + 100 s 2 + 5 s + 20 ) ) ) = − 5 + 100 t + 5 e − 20 t + ( 5 − 100 ( t − 2 ) − 5 e − 20 ( t − 2 ) u ( t − 2 ) ) − 200 ( 1 − e − 20 ( t − 2 ) ) u ( t − 2 ) Therefore, applying the definition of the Heaviside function the value of i(t) at the intervals 0 < t < 2 and t > 2 is obtained as follows. i ( t ) = ( − 5 + 100 t + 5 e − 20 t 0 < t < 2 e − 20 t ( 5 + e 40 ) t > 2 ) Thus, the value of i(t) is i ( t ) = ( − 5 + 100 t + 5 e − 20 t 0 < t < 2 e − 20 t ( 5 + e 40 ) t > 2 ) _ .

The value of the current, i ( t ) shown in the given circuit with the assumptions i ( 0 ) = 0 , R = 10 Ω and L = 0.5 H and using the Laplace transforms. The value of the current, i ( t ) is i ( t ) = ( − 5 + 100 t + 5 e − 20 t 0 < t < 2 e − 20 t ( 5 + e 40 ) t > 2 ) _ . Definition used: Laplace transform of derivatives: “The first and second derivatives of f ( t ) satisfy ℒ ( f ′ ) = s ℒ ( f ) − f ( 0 ) ⋯ ( 1 ) ℒ ( f ″ ) = s 2 ℒ ( f ) − s f ( 0 ) − f ′ ( 0 ) ⋯ ( 2 ) Formula (1) holds if f ( t ) is continuous for all t ≥ 0 and satisfies the growth restriction (2) and f ′ ( t ) is piecewise continuous on every finite interval on the semi axis t ≥ 0 . Similarly (2) holds if f ( t ) and f ′ ( t ) are continuous for all t ≥ 0 and satisfy the growth restriction and f ″ ( t ) is piecewise continuous on every finite interval on the semi axis t ≥ 0 .” Unit step function or Heaviside Function: The unit step function or the Heaviside function denoted by u ( t − a ) is defined as follows. u ( t − a ) = { 0 , if t < a 1 , if t > a } Second shifting theorem If f ( t ) has the transform F ( s ) , then the shifted function f ˜ ( t ) = f ( t − a ) u ( t − a ) = { 0 if t < a f ( t − a ) if t > a } has the transform e − a s F ( s ) . That is, if ℒ ( f ( t ) ) = F ( s ) then, ℒ ( f ( t − a ) u ( t − a ) ) = e − a s F ( s ) ⋯ ( 1 ) Calculation: Consider the given conditions as follows. i ( 0 ) = 0 , R = 10 Ω and L = 0.5 H Also, the value of voltage is given as shown below. v = { 200 t if 0 < t < 2 0 if t > 2 } Use the above definition of unit step function and rewrite the given voltage function in terms of unit step function as shown below. v ( t ) = 200 t ( 1 − u ( t − 2 ) ) = 200 t − 200 t u ( t − 2 ) It is known that the differential equation of a voltage of an RL circuit is L d I d t + R I = V ( t ) . Substitute the values R = 10 Ω and L = 0.5 H in the equation L d I d t + R I = V ( t ) as shown below. L d I d t + R I = V ( t ) 0.5 d I d t + 10 I = V ( t ) Take Laplace transform on the both sides of the above equation; apply definition of Laplace transform of derivatives as shown below. ℒ ( 0.5 d I d t + 10 I ) = ℒ ( V ( t ) ) 0.5 s ℒ ( I ) + 10 ℒ ( I ) = ℒ ( 200 t − 200 t u ( t − 2 ) ) ( 0.5 s + 10 ) ℒ ( I ) = 200 s 2 − 200 e − 2 s ( 1 s 2 + 2 s ) ( Multiply both sides by 2 ) ( s + 20 ) ℒ ( I ) = 2000 s 2 − 2000 ⋅ e − 2 s ( 1 + 2 s s 2 ) On further simplifications of the above equation the following is obtained. L ( I ) = 2000 s 2 ( s + 20 ) − 2000 ⋅ e − 2 s ( 1 + 2 s ) s 2 = 2000 s 2 ( s + 20 ) − 2000 e − 2 s s 2 ( s + 20 ) − 2000 e − 2 s ⋅ 2 s ( s + 20 ) = 2000 ( 1 − e − 2 s s 2 ( s + 20 ) − 2 e − 2 s s ( s + 20 ) ) = 2000 ( 1 − e − 2 s − 2 s e − 2 s s 2 ( s + 20 ) ) Now, apply partial fractions on the right hand side of the above equation as follows. 1 ( s 2 ) ( s + 20 ) = A s + B s 2 + C s + 20 1 ( s 2 ) ( s + 20 ) = A s ( s + 20 ) + B ( s + 20 ) + C s 2 ( s 2 ) ( s + 20 ) 1 = A s ( s + 20 ) + B ( s + 20 ) + C s 2 1 = A s 2 + 20 A s + B s + 20 B + C s 2 Equate the coefficient of s 2 on both sides of the above equation as follows. 0 = A + C A = − C Equate constant terms on both sides as follows. 1 = 20 B ( Divide both sides by 20 ) 1 20 = 20 B 20 1 20 = B Equate the coefficient of s on both sides of the above equation as follows. 0 = 20 A + B 0 = 20 A + 1 20 ( subtract 1 20 on both sides ) 0 − 1 20 = 20 A + 1 20 − 1 20 − 1 20 = 20 A ( Divide both sides by 20 ) − 1 400 = A Therefore, the value of C is 1 400 . Substitute the values of A , B and C in the equation 1 ( s 2 ) ( s + 20 ) = A s + B s 2 + C s + 20 as follows. 1 ( s 2 ) ( s + 20 ) = A s + B s 2 + C s + 20 = − 1 400 s + 1 20 s 2 + 1 400 ( s + 20 ) Thus, ℒ ( i ) = 2000 ( 1 − e − 2 s − 2 s e − 2 s ) ( − 1 400 s + 1 20 s 2 + 1 400 ( s + 20 ) ) . Rewrite the above as follows. ℒ ( i ) = 2000 ( 1 − e − 2 s − 2 s e − 2 s ) ( − 1 400 s + 1 20 s 2 + 1 400 ( s + 20 ) ) = ( 1 − e − 2 s − 2 s e − 2 s ) ( − 5 s + 100 s 2 + 5 s + 20 ) = ( − 5 s + 100 s 2 + 5 s + 20 ) − e − 2 s ( − 5 s + 100 s 2 + 5 s + 20 ) − 2 s e − 2 s ( − 5 s + 100 s 2 + 5 s + 20 ) Now, take inverse Laplace transform on both sides as follows. i ( t ) = ( ℒ − 1 ( − 5 s + 100 s 2 + 5 s + 20 ) − ℒ − 1 ( e − 2 s ( − 5 s + 100 s 2 + 5 s + 20 ) ) − 2 ℒ − 1 ( s e − 2 s ( − 5 s + 100 s 2 + 5 s + 20 ) ) ) = − 5 + 100 t + 5 e − 20 t + ( 5 − 100 ( t − 2 ) − 5 e − 20 ( t − 2 ) u ( t − 2 ) ) − 200 ( 1 − e − 20 ( t − 2 ) ) u ( t − 2 ) Therefore, applying the definition of the Heaviside function the value of i(t) at the intervals 0 < t < 2 and t > 2 is obtained as follows. i ( t ) = ( − 5 + 100 t + 5 e − 20 t 0 < t < 2 e − 20 t ( 5 + e 40 ) t > 2 ) Thus, the value of i(t) is i ( t ) = ( − 5 + 100 t + 5 e − 20 t 0 < t < 2 e − 20 t ( 5 + e 40 ) t > 2 ) _ .

ADVANCED ENGINEERING MATH W/ACCESS

ADVANCED ENGINEERING MATH W/ACCESS

10th Edition

ISBN: 9781119096023

Author: Kreyszig

Publisher: WILEY

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Chapter 6 Solutions

ADVANCED ENGINEERING MATH W/ACCESS

Ch. 6.1 - Prob. 1P Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....

Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Table 6.1. Convert this table to a table for...Ch. 6.1 - Using in Prob. 10, find , where f1(t) = 0 if t ≦...Ch. 6.1 - Table 6.1. Derive formula 6 from formulas 9 and...Ch. 6.1 - Nonexistence. Show that does not satisfy a...Ch. 6.1 - Nonexistence. Give simple examples of functions...Ch. 6.1 - Existence. Show that . [Use (30) in App. 3.1.]...Ch. 6.1 - Change of scale. If and c is any positive...Ch. 6.1 - Inverse transform. Prove that is linear. Hint:...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the shifted data IVPs by the Laplace...Ch. 6.2 - Solve the shifted data IVPs by the Laplace...Ch. 6.2 - Solve the shifted data IVPs by the Laplace...Ch. 6.2 - Solve the shifted data IVPs by the Laplace...Ch. 6.2 - Using (1) or (2), find if f(t) equals: t cos 4t Ch. 6.2 - Using (1) or (2), find if f(t) equals: te−at Ch. 6.2 - Using (1) or (2), find if f(t) equals: cos2 2t Ch. 6.2 - Using (1) or (2), find if f(t) equals: sin2 ωt Ch. 6.2 - Using (1) or (2), find if f(t) equals: sin4 t....Ch. 6.2 - Using (1) or (2), find if f(t) equals: cosh2 t Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.3 - Report on Shifting Theorems. Explain and compare...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Find and sketch or graph f(t) if equals e−3s/(s −...Ch. 6.3 - Prob. 13P Ch. 6.3 - Prob. 14P Ch. 6.3 - Find and sketch or graph f(t) if equals e−3s/s4 Ch. 6.3 - Prob. 16P Ch. 6.3 - Prob. 17P Ch. 6.3 - Using the Laplace transform and showing the...Ch. 6.3 - Using the Laplace transform and showing the...Ch. 6.3 - Prob. 20P Ch. 6.3 - Using the Laplace transform and showing the...Ch. 6.3 - Using the Laplace transform and showing the...Ch. 6.3 - Prob. 23P Ch. 6.3 - Using the Laplace transform and showing the...Ch. 6.3 - Prob. 25P Ch. 6.3 - Prob. 26P Ch. 6.3 - Prob. 27P Ch. 6.3 - Prob. 28P Ch. 6.3 - Prob. 29P Ch. 6.3 - Prob. 30P Ch. 6.3 - Prob. 31P Ch. 6.3 - Prob. 32P Ch. 6.3 - Prob. 33P Ch. 6.3 - Prob. 34P Ch. 6.3 - Prob. 35P Ch. 6.3 - Prob. 36P Ch. 6.3 - Prob. 37P Ch. 6.3 - Prob. 38P Ch. 6.3 - Prob. 39P Ch. 6.3 - Prob. 40P Ch. 6.4 - Prob. 3P Ch. 6.4 - Prob. 4P Ch. 6.4 - Prob. 5P Ch. 6.4 - Prob. 6P Ch. 6.4 - Prob. 7P Ch. 6.4 - Prob. 8P Ch. 6.4 - Prob. 9P Ch. 6.4 - Prob. 11P Ch. 6.4 - Prob. 15P Ch. 6.5 - CONVOLUTIONS BY INTEGRATION Find: Ch. 6.5 - CONVOLUTIONS BY INTEGRATION Find: 2. Ch. 6.5 - CONVOLUTIONS BY INTEGRATION Find: 3. Ch. 6.5 - CONVOLUTIONS BY INTEGRATION Find: 4. Ch. 6.5 - Prob. 5P Ch. 6.5 - Prob. 6P Ch. 6.5 - Prob. 7P Ch. 6.5 - Prob. 8P Ch. 6.5 - Prob. 9P Ch. 6.5 - Prob. 10P Ch. 6.5 - Prob. 11P Ch. 6.5 - Prob. 12P Ch. 6.5 - Prob. 13P Ch. 6.5 - Prob. 14P Ch. 6.5 - CAS EXPERIMENT. Variation of a Parameter. (a)...Ch. 6.5 - Prob. 17P Ch. 6.5 - Prob. 18P Ch. 6.5 - Prob. 19P Ch. 6.5 - Prob. 20P Ch. 6.5 - Prob. 21P Ch. 6.5 - Prob. 22P Ch. 6.5 - Prob. 23P Ch. 6.5 - Prob. 24P Ch. 6.5 - Prob. 25P Ch. 6.5 - Prob. 26P Ch. 6.6 - Prob. 2P Ch. 6.6 - Prob. 3P Ch. 6.6 - Prob. 4P Ch. 6.6 - Prob. 5P Ch. 6.6 - Prob. 6P Ch. 6.6 - Prob. 7P Ch. 6.6 - Prob. 8P Ch. 6.6 - Prob. 9P Ch. 6.6 - Prob. 10P Ch. 6.6 - Prob. 11P Ch. 6.6 - Prob. 14P Ch. 6.6 - Prob. 15P Ch. 6.6 - Prob. 16P Ch. 6.6 - Prob. 17P Ch. 6.6 - Prob. 18P Ch. 6.6 - Prob. 19P Ch. 6.6 - Prob. 20P Ch. 6.7 - Prob. 2P Ch. 6.7 - Prob. 3P Ch. 6.7 - Prob. 4P Ch. 6.7 - Prob. 5P Ch. 6.7 - Prob. 6P Ch. 6.7 - Prob. 7P Ch. 6.7 - Prob. 8P Ch. 6.7 - Prob. 9P Ch. 6.7 - Prob. 10P Ch. 6.7 - Prob. 11P Ch. 6.7 - Prob. 12P Ch. 6.7 - Prob. 13P Ch. 6.7 - Prob. 14P Ch. 6.7 - Prob. 15P Ch. 6.7 - Prob. 16P Ch. 6.7 - Prob. 19P Ch. 6.7 - Prob. 20P Ch. 6 - Prob. 1RQ Ch. 6 - Prob. 2RQ Ch. 6 - Prob. 3RQ Ch. 6 - Prob. 4RQ Ch. 6 - Prob. 5RQ Ch. 6 - When and how do you use the unit step function and...Ch. 6 - If you know f(t) = ℒ−1{F(s)}, how would you find...Ch. 6 - Explain the use of the two shifting theorems from...Ch. 6 - Prob. 9RQ Ch. 6 - Prob. 10RQ Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the inverse transform, indicating the method...Ch. 6 - Prob. 21RQ Ch. 6 - Prob. 22RQ Ch. 6 - Prob. 23RQ Ch. 6 - Prob. 24RQ Ch. 6 - Prob. 25RQ Ch. 6 - Prob. 26RQ Ch. 6 - Prob. 27RQ Ch. 6 - Prob. 28RQ Ch. 6 - Prob. 29RQ Ch. 6 - Prob. 30RQ Ch. 6 - Prob. 31RQ Ch. 6 - Prob. 32RQ Ch. 6 - Prob. 33RQ Ch. 6 - Prob. 34RQ Ch. 6 - Prob. 35RQ Ch. 6 - Prob. 36RQ Ch. 6 - Prob. 37RQ Ch. 6 - Prob. 38RQ Ch. 6 - Prob. 39RQ Ch. 6 - Prob. 40RQ Ch. 6 - Prob. 41RQ Ch. 6 - Prob. 42RQ Ch. 6 - Prob. 43RQ Ch. 6 - Prob. 44RQ Ch. 6 - Prob. 45RQ

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