6(2t)=limb△(2t) The area under which remains 3. From the result in Section 1.4.2 of the text we 6(2)=亓6(t) We can also show this using the relationship between the unit step function and the unit impulse (i.e. the unit impulse is the time derivative of the unit step function). For a given A, the approximation of both unit steps t) and uA(2t)are shown to the right. Note that u,△(2t) reaches unity at t=会 Making a change of variable s= 2t where ds 6△(2)6△(t) 2dt and using Eqn(1. 73)in O& w, gives 2dt 1du△(2)12 2 dt 2△△ The area enclosed by a(2t) is half of that of Sa(t) as shown to the right. Since &(t)is defined by it's area, we get d(2t)=38(t). In general 6(a)=一0(t) holds for any nonzero real number a. The above is a proof of the time scaling property, which tells us that we've simply squeezed da(t) by a� � �(2t) = lim ��(2t) ��0 The area under which remains 1 . From the result in Section 1.4.2 of the text we 2 have: 1 �(2t) = �(t) 2 u�(2t) We can also show this using the relationship 1 between the unit step function and the unit �� u�(t) impulse (i.e. the unit impulse is the time � derivative of the unit step function). For a given �, t the approximation of both unit steps � 0 � u�(t) and u�(2t) are shown to the right. Note 2 � that u�(2t) reaches unity at t = �. 2 Making a change of variable s = 2t where ds = ��(2t) ��(t) 1 2dt and using Eqn(1.73) in O&W, gives � du�(s) du�(2t) ��(s) = = ds 2dt 1 du�(2t) 1 2 1 = = = . 2 dt 2 � � t 0 � The area enclosed by ��(2t) is half of that of 2 � ��(t) as shown to the right. Since �(t) is defined by it’s area, we get �(2t) = 1 �(t). In general 2 1 �(at) = �(t) |a| holds for any nonzero real number a. The above is a proof of the time scaling property, which tells us that we’ve simply squeezed ��(t) by a factor of two. 2