Question 4 determine the inverse laplace transform of the following signals by computing the partial fraction expansion analytically for each case. then check your answer by using the matlab function residue to find the poles and residues (the coefficients ci of the partial fraction expansion). type residue in matlab command window for details on how to use residue. hint: write x(s) in (b) as 1 + a strictly proper fraction. check your result x(t) by using the matlab function ilaplace. plot the x(t) for each case using the matlab function ezplot in the range [-1 to 10]. do you see anything unexpected when you compare the plots for (a) and (b)? try to plot the difference of the two x(t)’s and check whether matlab can plot it properly, and then also comment. x(s)=(s+1)/(〖s^2+ 5s〗^ +6) x(s)=(〖s^2+ 6s〗^ +7)/(〖s^2+ 5s〗^ +6)

bull

step-by-step explanation:

the degrees would remain the same ( the number to the left of the decimal point.

the multiply the decimal by 60 to get the minutes and use the numbers to the left of the decimal point.

then multiply the decimal portion of that by 60 for the seconds and round off as needed.

39 .956652

the 39 remains the degrees.

the minutes = 0.956652 x 60 = 57.39912

the minutes = 57

the seconds = 0.39912 x 60 = 23.95 = 24 ( rounded to nearest tenth)

so 39.956652 = 39 degrees, 57 minutes and 24 seconds

-75.159809

the -75 remains the degrees

the minutes = 0.159809 x 60 = 9.58854

the minutes = 9

the seconds = 0.58854 x 60 = 35.31, rounded to nearest tenth = 35.3 seconds

so -75.159809 = -75 degrees, 9 minutes and 35.3 seconds.

the first coordinate is north or south, since it it positive it is north.

the second number is east or west, because it is negative it is west.

39 degrees, 57 minutes and 24 seconds n

-75 degrees, 9 minutes and 35.3 seconds w

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