EECE 301 Signals & Systems Prof. Mark FowlerEECE 301 Signals & SystemsProf. Mark FowlerDiscussion #3a• Review of Differential EquationsDifferential Equations ReviewDifferential Equations like this are Linear and Time Invariant:)()(...)()(...)()(010111tfbdttdfbdttfdbtyadttydadttydammmnnnnnn+++=+++−−−-coefficients are constants ⇒ TI-No nonlinear terms ⇒ Linear.,)()(),(,)()(),( etcdttyddttydtydttyddttydtfppkknppkkn⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡Examples of Nonlinear Terms:In the following we will BRIEFLY review the basics of solving Linear, Constant Coefficient Differential Equations under the HomogeneousCondition“Homogeneous” means the “forcing function” is zeroThat means we are finding the “zero-input response” that occurs due to the effect of the initial coniditions.⇒ Write D.E. like this:()())(...)(...)(01)(0111tfbDbDbtyaDaDaDDPmmDQnnn4443444214444434444421ΔΔ==−−+++=++++⇒.. EqDiff)()()()( tfDPtyDQ=m is the highest-order derivative on the “input” siden is the highest-order derivative on the “output” sideWe will assume: m ≤ n)()(tyDdttydkkk≡Use “operational notation”:Due to linearity: Total Response = Zero-Input Response + Zero-State ResponseZ-I Response: found assuming the input f(t) = 0 but with given IC’sZ-S Response: found assuming IC’s = 0 but with given f(t) applied()00)(...0)()(:..0111>∀=++++⇒=⇒−−ttyaDaDaDtyDQEDzinnnzi(▲)numberscomplex possibly areand)(Consider 0λλccetyt=“linear combination” of yzi(t) & its derivatives must be = 0Can we find c and λ such that y0(t) qualifies as a homogeneous solution? Finding the Zero-Input Response (Homogeneous Solution) Assume f(t) = 0Put y0(t) into (▲) and use result for the derivative of an exponential:0)...(0111=++++−−tnnneaaacλλλλmust = 0 solutiona is solutiona is solutiona is 2121tnttnecececλλλMThen, choose c1, c2,…,cnto satisfy the given IC’stnttzinececectyλλλ+++= ...)(:Solution I-Z2121tnntnedtedλλλ=Characteristic polynomialQ(λ) has at most n unique roots (can be complex)))...()(()(21 nQλλλλλλλ−−−=⇒So…any linear combination is also a solution to (▲){}nitie1=λSet of characteristic modesReal Root: tiiiejσσλ⇒+= 0real real t0>iσ0=iσ0<iσiiijωσλ+=tjttiiieeeωσλ+=0<iσt0>iσt0=iσtComplex Root:Mode:To get only real-valued solutions requires the system coefficients to be real-valued. ⇒ Complex roots of C.E. will appear in conjugate pairs:ωσλωσλjjki−=+=Conjugate pairtjtktjtitktieeceecececkiωσωσλλ+=+For some real CθjeC2θjeC−20)cos( >+ ttCetθωσUse Euler!Repeated RootsSay there are r repeated roots))...()(()()(321 prQλλλλλλλλλ−−−−=p = n - rZI Solution: ():modesother ...)(111211++++=−trrziietctcctyλeffect of r-repeated rootsSee examples on the next several pagesWe “can verify” that:trtttetetteei11112,...,,,λλλλ−satisfy (▲)Find the zero-input response (i.e., homogeneous solution) for these three Differential Equations.Example (a))()(2)(3)()()(2)(3)(222tDftytDytyDdttdftydttdydttyd=++=++The zero-input form is:0)(2)(3)(0)(2)(3)(222=++=++tytDytyDtydttdydttydThe Characteristic Equation is:0)2)(1(0232=++⇒=++λλλλDifferential Equation Examplesw/ I.C.’sThe Characteristic Equation is:0)2)(1(0232=++⇒=++λλλλ2&121The Characteristic Roots are:−=−=λλThe Characteristic “Modes” are:tttteeee221&−−==λλThe zero-input solution is:ttzieCeCty221)(−−+=The System forces this form through its Char. Eq.The IC’s determine the specific values of the Ci’sThe zero-input solution is:ttzieCeCty221)(−−+=and it must satisfy the ICs so:0)0(0210201=+⇒+==−−CCeCeCyziThe derivative of the z-s soln. must also satisfy the ICs so:522)0(5210201=+⇒−−=′=−−−CCeCeCyziTwo Equations in Two Unknowns leads to:5&521=−=CCThe “particular” zero-input solution is:321321mode second2modefirst 55)(ttzieety−−+−=Because the characteristic roots are real and negative…the modes and the Z-I response all decay to zero w/o oscillations 0 1 2 3 4 5 6 7 8 9 10-6-4-20First ModePlots for Example 2.1 (a)0 1 2 3 4 5 6 7 8 9 100246Second Mode0 1 2 3 4 5 6 7 8 9 10-1.5-1-0.50Zero-Input Responset (sec)Plots for Example (a)slower decay for the e-ttermfaster decay for the e-2ttermExample (b):)(5)(3)(9)(6)()(5)(3)(9)(6)(222tftDftytDytyDtfdttdftydttdydttyd+=+++=++The zero-input form is:0)(9)(6)(0)(9)(6)(222=++=++tytDytyDtydttdydttydThe Characteristic Equation is:0)3(09622=+⇒=++λλλw/ I.C.’sThe Characteristic Equation is:The Characteristic Roots are:3&321−=−=λλThe Characteristic “Modes” are:ttttteteee3321&−−==λλThe zero-input solution is:ttziteCeCty3231)(−−+=The System forces this form through its Char. Eq.The IC’s determine the specific values of the Ci’s0)3(09622=+⇒=++λλλUsing the “rule” to handle repeated rootsFollowing the same procedure (do it for yourself!!) you get…The “particular” zero-input solution is:tttzietteety3mode second3modefirst 3)23(23)(−−−+=+=3213210 1 2 3 4 5 6 7 8 9 100123First ModePlots for Example 2.1 (b)0 1 2 3 4 5 6 7 8 9 1000.20.4Second Mode0 1 2 3 4 5 6 7 8 9 100123t (sec)Zero-Input ResponsePlots for Example (b)Same Decay RatesEffect of “t out in front”Because the characteristic roots are real and negative…the modes and the Z-I response all .Example (c):)(2)()(40)(4)()(2)()(40)(4)(222tftDftytDytyDtfdttdftydttdydttyd+=+++=++The zero-input form is:0)(40)(4)(0)(40)(4)(222=++=++tytDytyDtydttdydttydThe Characteristic Equation is:0)62)(62(04042=++−+⇒=++ jjλλλλw/ I.C.’sThe Characteristic Equation is:The Characteristic Roots are:62&6221jj−−=+−=λλThe Characteristic “Modes” are:tjtttjtteeeeee626221&−−+−==λλThe zero-input solution is:tjttjtzieeCeeCty622621)(−−+−+=The System forces this form through its Char. Eq.The IC’s determine the specific values of the Ci’s0)62)(62(04042=++−+⇒=++ jjλλλλFollowing the same procedure with some manipulation of complex exponentials into a cosine…The “particular” zero-input solution is:)3/6cos(4)(2π+=−tetytziSet by the ICsImag. part of root controls oscillationReal part of root controls Decay0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 100.51First ModePlots for Example 2.1 (c)0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 100.51Second Mode0 0.5 1 1.5 2 2.5 3 3.5 4-4-2024t (sec)Zero-Input ResponseCan’t Easily Plot the Modes Because They Are ComplexPlots for