(Test Bank) Contemporary Communication Systems

Mesiya, 1

Engineering

Test Bank

1st Edition

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Test Bank for Contemporary Communication Systems, The downloadable file consists of 398 pages, and the total size of the file is 3.77 MB. Purchase now only for highly discounted price.

Author(s): Mesiya

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1 Chapter 2 2.1 Consider the signals displayed in Figure P2.1.Show that each of these signals can be expressed as the sum of rectangular ()tΠ and/or triangular ()tΛ pulses. Figure P2.1 Solution: a. 1()24ttxt⎛⎞ ⎛⎞=Π +Π⎜⎟ ⎜⎟⎝⎠ ⎝⎠ b. 233() 262ttxt−−⎛⎞ ⎛⎞=Λ − Λ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ c. 311 7 3 1 5 9( ) ... ...2 22222x tt ttttt⎛ ⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞=+ Π+ + Π+ + Π+ + Π− + Π− + Π− +⎜ ⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎝ ⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ()20 . 5 ntn∞ =−∞=Π −+⎡⎤⎣⎦∑ © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 2d. 444( ) ... ...222tt txt+−⎛⎞ ⎛ ⎞ ⎛⎞=+ Λ + Λ + Λ +⎜⎟ ⎜ ⎟ ⎜⎟⎝⎠ ⎝ ⎠ ⎝⎠ 4 2 ntn∞ =−∞−⎛⎞=Λ⎜⎟⎝⎠∑ 2.2 For the signal 2()xt in Figure P2.1 (b) plot the following signals a. 2(3 )xt− b. 2()xt− c. 2(2 )xt d. 2(3 2 )x t− Solution: 2.3 Plot the following signals a. 1() 2 ( / 2 ) c o s ( 6 )xtt t π =Π b. ()211() 2 s g n22xtt⎡⎤=+⎢⎥⎣⎦ t2()xt− 1 0 3 122(3 )xt− 1 40 3 12 56 9 784− 6− 3−5− 2−1− t2(2 )xt 1 3 1 40 2 56t2(3 2 )xt− 1 3 1 40 2 2−1−t2()xt− 1 0 3 122(3 )xt− 1 40 3 12 56 9 782(3 )xt− 1 40 3 123 12 56 9 784− 6− 3−5− 2−1− t2(2 )xt 1 3 1 40 2 56t2(2 )xt 1 3 1 40 2 56t2(3 2 )xt− 1 3 1 40 2 2−1−t2(3 2 )xt− 1 3 1 40 2 2−1− © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 3c. 32() ( 2 )xt x t=− + d. ()( )4() s i n c 2 / 2xtt t=Π Solution: 2.4 Determine whether the following signa ls are periodic. For periodic signals, determine the fundamental period. a. 1( ) sin( ) 5cos(4 / 5)xtt tπ π =+ Solution: sin( ) tπis periodic with period 122 Tπ π==. cos(4 / 5) tπ is periodic with period 225 4/ 5 2Tπ π== . 1()xt is periodic if the ratio 1 2T Tcan be written as ratio of integers. In the present case, 1 222 4 55T T×== -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-0.4-0.200.20.40.60.81 t t3()xt 2 3 1 40 2" t-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-2-1.5-1-0.500.511.52 ttt2()xt 2 3 1 40 2 56" -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-0.4-0.200.20.40.60.81 t t3()xt 2 3 1 40 2" t-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-2-1.5-1-0.500.511.52 ttt2()xt 2 3 1 40 2 56" t2()xt 2 3 1 40 2 56" © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 4 Therefore, 1()xt is periodic with fundamental period oTsuch that 12541 0oTTT=== b. 39 2() c o s ( 1 2)jt jtxte e t=++ Solution: 3jteis periodic with period 12 3Tπ= . Similarly, 9jteis periodic with period 22 9Tπ= . cos(12 ) t is periodic with period 32 12 6Tππ==. 2()xtis periodic with fundamental period ()1232,,3oTL C M T T Tπ== . c. 3( ) sin(2 ) cos(10 )xtt tπ=+ Solution: sin(2 ) tπis periodic with period 1212Tπ π==. cos(10 ) t is periodic with period 22 10 5Tππ== . 3()xt is periodic if the ratio 1 2T Tcan be written as ratio of integers. In the present case, 1 215 5 T Tππ×== Since π is an irrational number, the ratio is not rational. Therefore, 3()xtis not periodic. d. 4() c o s 2 s i n ( 5 )4xtt tππ π⎛⎞=− +⎜⎟⎝⎠ Solution: cos 24tππ⎛⎞−⎜⎟⎝⎠ is periodic with period 1212Tπ π==. sin(5 ) tπis periodic with period 222 55Tπ π==. 4()xtis periodic if the ratio 1 2T Tcan be written as ratio of integers. In the present case, © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 5 1 215 5 22T T×== Therefore, 4()xtis periodic with fundamental period oTsuch that 12252oTT T=== 2.5 Classify the following signals as odd or even or neither. a. ( ) 4xtt=− Solution: ()4 ( 4 ) ( ) xtt t x t−= = − − = − . So ( ) xtis odd. b. ()txte−= Solution: ttee−− −= . So ()xtis even. c. ( ) 5cos(3 )xtt= Solution: Since cos( ) tis even, 5cos(3 ) tis also even. d. () s i n ( 3 )2xt tπ=− Solution: ( ) sin(3 ) cos(3 )2xtt tπ=− = − which is even. e. () ()xtu t= Solution: ()ut is neither even nor odd; For example, (1) 1 but ( 1) 0 (1)uu u=−=≠ − . © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 6f. ( ) sin(2 ) cos(2 )xttt=+ Solution: ( )xt is neither even nor odd; For example, ( / 8) 2 but ( / 8) 0 ( / 8)xx xππ π=− = ≠ − . 2.6 Determine whether the following signals are energy or power, or neither and calculate the corresponding ener gy or power in the signal: a. 1() ()xtu t= Solution: The normalized average power of a signal 1()xtis defined as 1/2 /2 2 /2 011 1 1lim ( ) lim 1 lim22TT xTT T TTPu t d t d tTT T→∞ →∞ →∞ −== = =∫∫ Therefore, 1()xt is a power signal. b. ()2() 4 c o s ( 2 ) 3 c o s 4xttt π π =+ Solution: 4cos(2 ) tπ is a power signal. () 3cos 4 tπis also a power signal. Since 2()xtis sum of two power signals, it is a power signal. () () () ()2/2 /222 /2 /2 /2 22 /211lim 4cos(2 ) 3cos 4 lim 4cos(2 ) 3cos 4 1 lim 16cos (2 ) 9cos 4 24cos(2 )cos 4TT xTT TT T T TP t td t t td tTT ttt t d tTππ ππ πππ π→∞ →∞ −− →∞ −⎡⎤ ⎡⎤ =+ =+⎣⎦ ⎣⎦ ⎡⎤ =+ +⎣⎦∫∫ ∫ Now []/2 /2 2 /2 /2 /2 2 /218 8lim 16cos (2 ) lim 1 cos(4 ) lim 8 1lim 9cos (4 ) 4.5TT TT T TT T T TTtd t t d tTT T td tTππ π→∞ →∞ →∞ −− →∞ −= += = =∫∫ ∫ © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 7 () ()/2 /2 /2 /224 24lim cos(2 )cos 4 lim cos(6 ) cos 2 0TT TT TTt t dt t t dtTTππ π π →∞ →∞ −−⎡⎤ = +=⎣⎦ ∫∫ Therefore, 28 4.5 12.5xP=+ = c. 31()xtt= Solution: 3/2 /2 /2 2 2 /2 /2 /212lim 1/ lim lim lim 0/2T TT xTT T TT TTEt dt t d ttT− →∞ →∞ →∞ →∞− −−⎛⎞ −⎛⎞== = − = = ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠∫∫ 3/2 /2 /2 2 2 /2 /2 /211 1 1 1 2lim 1/ lim lim lim 0/2T TT xTT T TT TTPt dt t dtTT T t T T− →∞ →∞ →∞ →∞− −−⎛⎞ −⎛⎞== = − = = ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠∫∫ 3()xt is neither an energy nor a power signal. d. 4() ()txte u tα−= Solution: ()4/2/2 /2 222 /2 0 011lim ( ) lim lim lim 122 2TTT t tt T xTT T T TeEe u t d t e d t eα αα α α αα− −− − →∞ →∞ →∞ →∞ −⎛⎞ ⎜⎟ == = − = − + =⎜⎟⎝⎠∫∫ Thus 4()xt is an energy signal. e. 5() ( / 3 ) ()xtt t=Π +Π Solution: t3/21 3/2−25()xt t3/21 3/2−25()xt © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 8 5/2 1/2 1/2 1/2 2 /2 3/2 1/2 3/2lim ( / 3) ( ) 1 4 1 1 4 1 6T xT TEt t d t d t d t d t−− →∞ −− − −=Π + Π = + + =++=∫∫ ∫ ∫ Thus 5()xt is an energy signal. f. (2 1 0 ) 6() 5 ()tj txte u tπ−+= Solution: ()6/2 /2 /222( 2 10 ) 2 10 4 /2 0 0 /24 2 0lim 5 ( ) lim 5 ) lim 25 25 25 25 lim lim 144 4TT T tj t t j t t xTT T T Tt T TTE e u t dt e e dt e dt eeππ−+ − − →∞ →∞ →∞ − − − →∞ →∞== = ⎛⎞ ⎜⎟=−=− + =⎜⎟⎝⎠∫∫ ∫ Thus 6()xt is an energy signal. g. ()7() 4 / 2 nxt t n∞ =−∞=Λ −⎡⎤⎣⎦∑ Solution: 7()xt is a periodic signal with period 4oT=. () () ()7/2 112 2 2 /2 0 0 13 2 012 1/2 1 1 22 11 1 1 1 123 2 3 6o oT x ooTP t d t td t ttd tTT ttt−⎡⎤=Λ = − = − +⎣⎦ ⎛⎞ ⎛⎞=− + =− + =⎜⎟ ⎜⎟⎝⎠ ⎝⎠∫∫ ∫ 2.7 Evaluate the following expressions by us ing the properties of the delta function: a. 1( ) (4 )sin(2 )xtt tδ= Solution: 1(4 ) ( )4ttδδ= 111() () s i n ( 2) () s i n ( 0 ) 044xt t t tδδ== = © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 9b. 2() () c o s 3 04xt t tπδπ⎛⎞=+⎜⎟⎝⎠ Solution: 2() () c o s 3 0 () c o s 0 c o s ()44 4xtt t t tππ πδ πδ δ⎛⎞ ⎛ ⎞ ⎛ ⎞=+ = + =⎜⎟ ⎜ ⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ⎝ ⎠ c. 3() () s i n c ( 1 )xt t tδ=+ Solution: 3( )( ) s i n c ( 1 )( ) s i n c ( 0 1 )( ) s i n c ( 1 )( ) 0 0xt t t t t tδ δδ δ =+ =+ = = × = d. 4( ) ( 2) sin(2.5 )txtt e tδ π−=− Solution: 4 22() ( 2 ) s i n ( 2 . 5 ) ( 2) sin(2.5 2) ( 2) sin( ) 0txt t e t te teδπ δπ δ π− −−=− =− × =− = e. 5() ( 2) s i n c ()xtt t d tδ∞ −∞=∫ Solution: 51( ) (2 )sinc( ) ( )sinc( )2 11 1 ( )sinc(0) ( )22 2xt t td t t td t td t t d tδδ δδ∞∞ −∞ −∞ ∞∞ −∞ −∞== == =∫∫ ∫∫ f. 6() ( 3 ) c o s ()xtt t d tδ∞ −∞=−∫ Solution: 6() ( 3 ) c o s () c o s ( 3 ) ( 3 ) c o s ( 3 )xt t t d t t d tδδ∞∞ −∞ −∞=− = − =∫∫ © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 10g. 7 31() ( 2 )1xtt d ttδ∞ −∞=−−∫ Solution: 7 33 311() ( 2 ) ( 2 )11 11 1 ( 2) ( 2)12 7 7xt t d t t d ttt td tt d tδδ δδ∞∞ −∞ −∞ ∞∞ −∞ −∞=− =−−− =− = − − = −−∫∫ ∫∫ h. 3 8() ( 3 4 )txtt e d tδ∞ − −∞=−∫ Solution: 33 8 4 443314() ( 3 4 )33 14 4 33 3 3 3ttx t t e dt t e dt eete d t td tδδ δδ∞∞ −− −∞ −∞ ∞∞ − −−× −∞ −∞⎛⎞=− = − ⎜⎟⎝⎠ ⎛⎞ ⎛⎞=− = − =⎜⎟ ⎜⎟⎝⎠ ⎝⎠∫∫ ∫∫ i. 9() ' () ()xtttδ=⊗ Π Solution: () ()9 0 '( ) () ' ( ) ( 1 ) () ( ) 0.5 0.5dxt t d td tt tττδ τ τ ττ δδ∞ = −∞=Π− = − Π− =Π = + − −∫ 2.8 For each of the following continuous-time system s, determine whether or not the system is (1) linear, (2) time-invariant, (3) memoryless, and (4) casual. a. () ( 1 )yt xt=− Solution: The system is linear, time-invariant, causal, and has memory. The system has memory because current value of the out put depends on the previous value of the input. The system is causal becaus e current value of the output does not depend on future inputs. To prove linearity, let 12 () () ()xtx tx tαβ =+ . The response of the system to ()xt is © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 11 12 12( ) (1 ) (1 ) (1 ) ( ) ( )yt xt x t x t yt ytα β αβ=− = − + − =+ where 11 2 2() () a n d () ()xty t x ty t⎯⎯→⎯ ⎯→TT. To show that the system is time-invariant, let ) ( ) (1 ot t x t x−= be the system input. The corresponding output is 11() ( 1 ) ( 1 ) ( )oo yt xt x t t y t t=− =− − =− b. ( ) 3 ( ) 2yt xt=− Solution: The system is nonlinear, time-invariant, causal, and memoryless. The system is memoryless because current value of the output depends only on the current value of the input. The system is causal because current value of the output does not depend on future inputs. To prove nonlinearity, let 12 () () ()xtx tx tαβ =+ . The response of the system to ( )xt is [ ] [] []12 12 1 2() 3 () 2 3 () () 2 ( ) ( ) 3 ( ) 2 3 ( ) 2yt xt x t x t yt yt xt xtαβ αβα β=− = + − ≠+= − + − To show that the system is time-invariant, let ) ( ) (1 ot t x t x−= be the system input. The corresponding output is 11( )3 ( )23 ( )2 ( )oo yt xt x t t y t t=− = − − = − c. () ()yt xt= Solution: The system is nonlinear, time-invariant, causal, and memoryless. The system is memoryless because current value of the output depends only on the current value of the input. The system is causal because current value of the output does not depend on future inputs. To prove nonlinearity, let 12 () () ()xtx tx tαβ =+ . The response of the system to ()xt is © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 12 12 1 2 () () () () ()y t xt xt xt xtαβ α β=+≤+ Because 12 1 21 2() () () () () () f o r a l l a n d xt xt xt xt yt y tαβα β α β α β+= + ≠ + , the system is nonlinear. To show that the system is time-invariant, let ) ( ) ( 1 ot t x t x−= be the system input. The corresponding output is 11() () ( ) ( )oo yt xt x t t y t t== − = − d. [] () c o s ( 2) ()yt t xt= Solution: The system is linear, time-variant, causal, and memoryless. The system is memoryless because current value of the output depends only on the current value of the input. The system is causal because current value of the output does not depend on future inputs. To prove nonlinearity, let 12 () () ()xtx tx tαβ =+ . The response of the system to()xt is [][][ ] [] []12 12 1 2() c o s ( 2) () c o s ( 2) () () c o s ( 2) () c o s ( 2) () () ()yt t xt t x t x t tx t tx t y t y tαβ αβα β== + =+= + To prove time-variance, let ) ( ) (1 ot t x t x−= be the system input. The corresponding output is [][]() ()( )11() c o s ( 2) () c o s ( 2) ( ) c o s 2oo o o y t t xt t xt t y t t t t xt t== − ≠ − = − − ⎡⎤⎣⎦ e. ()()xtyt e= Solution: The system is nonlinear, time-invariant, causal, and memoryless. The system is memoryless because current value of th e output depends only on the current value of the input. The system is causal because current value of the output does not depend on future inputs. © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 13 To prove nonlinearity, let 12 () () ()xtx tx tαβ =+ . The response of the system to ( )xt is 12 1 2 1 2 () () () () () () 12 () () ()xt x t xt x t xt x tyt e e e y t y t e eαβ αβαβαβ+==≠ + = + for all and αβ To show that the system is time-invariant, let ) ( ) (1 ot t x t x−= be the system input. The corresponding output is 1 () () 1() ( )o xt t xt o yt e e y t t−== = − f. ( ) ( ) yt t xt= Solution: The system is linear, time-variant, causal, and memoryless. The system is memoryless because current value of th e output depends only on the current value of the input. The system is causal because current value of the output does not depend on future inputs. To prove nonlinearity, let 12 () () ()xtx tx tαβ =+ . The response of the system to ( ) xt is [ ] 12 1 2 12 () () () () () () ()yt t x t x t t x t t x t y t y tαβ αβ αβ=+ = + = + To prove time-variance, let ) ( ) (1 ot t x t x−= be the system input. The corresponding output is () ()()11() () ( )oo o o y t t xt t xt t y t t t t xt t== − ≠ − = − − g. ( ) (2 )t yt x d ττ −∞=∫ Solution: The system is linear, time-invariant, causal, and has memory. The system has memory because current value of the outp ut depends only on the past values of the input. The system is causal becaus e current value of the output does not depend on future inputs. To prove linearity, let 12 () () ()xtx tx tαβ =+ . The response of the system to ( ) xt is © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 14[] 12 1 2 1 2 ( ) (2 ) (2 ) (2 ) (2 ) ( ) ( )tt t yt x x d x d x d y t y tα τ β τ τα ττβ ττα β −∞ −∞ −∞=+ = + = +∫∫ ∫ To check for time-variance, let ) ( ) (1 ot t x t x−= be the system input. The corresponding output is [] () ( )11() ( 2 ) 2 ( ) 2ott tt oo yt x d x t d x d yt t ττ τ τ αα− −∞ −∞ −∞== − = = −∫∫ ∫ 2.9 Calculate the output ()yt of the LTI system for the following cases: a. 2( ) ( ) and ( ) ( 2) ( 4)txt e ut ht ut ut−== −−− Solution: 3() ( 2 ) ( 4 )2tht ut ut−⎛⎞=− −− = Π ⎜⎟⎝⎠ 2( ) 3( ) ( ) () ()2tyt h xt d e ut dτ ττττ τ τ∞∞ −− −∞ −∞−⎛⎞=− = − Π ⎜⎟⎝⎠∫∫ For 2t<, there is no overlap and ( ) 0. yt= For 24t≤< , 22 22 ( 2 ) 2( ) 2 20 01()22ttt t t eeyt e d e dτ ττττ−− −− − −− − −== = =− ∫∫ For 4t≥, ()4 4 44 28 4 2( ) 2 2 2 2 2( 2) 22 21 ()22 2tt t t te ee eyt e d e e d e e eτ ττττ−− − − − −−− −=== ==∫∫ ()2( 2) 4 2( 2)1, 2 42 1 () , 42 0, otherwiset tet e yt e t−− −−⎧−≤≤ ⎪ ⎪⎪ − ⎪=>⎨⎪ ⎪ ⎪ ⎪⎩ b. 2() () a n d () ()ttxte u t h te u t−−== © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 15Solution: 2( ) 2( ) 0() ( ) ( )t ttyt e u e ut d e e dττ τ ττ ττ τ∞ −− − − − − −∞=− = ∫∫ For 0t<, there is no overlap and ( ) 0. yt= ()22 2 2 0 0() 1 , 0tttt t t t tyt e ed e e e e e e tτττ−− − − −== = − = − ≥∫ c. 2() ( ) a n d () () 3 ()txtu t h t t e u t δ−=− = − Solution: 2( ) ()( ) () 3 () ( )yt h xt d e u u tdττττ δ τ τ τ τ∞∞ − −∞ −∞⎡⎤ =− = − −⎣⎦ ∫∫ Now ()( ) ( ) ut d u t δτ τ τ∞ −∞−= −∫ For 0t<, 2 2 0 033322eedτ ττ∞∞ − −==−∫ For 0t≥, 2 22 2 333( ) ( ) 3 22t t teeu u t d ed eτ ττττ τ τ∞∞∞ − −− − −∞−= = =− ∫∫ 2351 , 022()3, 02tt yt et−⎧+= <⎪⎪=⎨ ⎪ ≥⎪⎩ d. 3() ( 2 ) 3 ( ) a n d () () ( 1 )txt t e u t ht ut utδ=− + − =−− Solution: © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.