4 added 648 characters in body edited Jan 16 at 16:00 Matt L. 52.4k22 gold badges3939 silver badges9898 bronze badges If the problem description is correct, i.e., if $$h[n]$$ is zero for $$n<0$$ and $$n>7$$, and if we assume that $$h[7]\neq 0$$ (and why should we assume otherwise?), then the filter has $$8$$ taps and it is a $$7^{th}$$ order FIR filter with $$7$$ zeros. Since it is a linear phase filter and the number of taps is even, it must be either a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $$z=-1$$, in the second case it must have a zero at $$z=1$$. Refer to this answer for more details on the $$4$$ types of linear phase FIR filters. So if we assume even symmetry (type II), the zeros are $$z_0=0.4e^{j\pi/3}$$, $$z_1=z_0^*$$, $$z_2=1/z_0$$, $$z_3=1/z_0^*$$, $$z_4=3$$, $$z_5=1/z_4$$, $$z_6=-1$$. If we assume odd symmetry (type IV), the zeros are $$z_0=0.4e^{j\pi/3}$$, $$z_1=z_0^*$$, $$z_2=1/z_0$$, $$z_3=1/z_0^*$$, $$z_4=3$$, $$z_5=1/z_4$$, $$z_6=1$$. The corresponding impulse responses are $$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$ $$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$ EDIT: The answer above is correct (as far as I can tell) for the given problem description. Now that I've checked the original source (Schaum's Outlines of DSP), which includes the solution, I believe that there are the following possibilities: either there's a typo in the problem description and they actually meant that $$h[n]$$ is zero for $$n<0$$ and $$n\ge 7$$ (note the "greater or equal" sign). In this case we really have a $$6^{th}$$ order FIR filter and the given solution in Schaum's Outline is correct. The other option would be that they don't know what they're talking about (and I hope and believe that this is not the case). If the problem description is correct, i.e., if $$h[n]$$ is zero for $$n<0$$ and $$n>7$$, and if we assume that $$h[7]\neq 0$$ (and why should we assume otherwise?), then the filter has $$8$$ taps and it is a $$7^{th}$$ order FIR filter with $$7$$ zeros. Since it is a linear phase filter and the number of taps is even, it must be either a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $$z=-1$$, in the second case it must have a zero at $$z=1$$. Refer to this answer for more details on the $$4$$ types of linear phase FIR filters. So if we assume even symmetry (type II), the zeros are $$z_0=0.4e^{j\pi/3}$$, $$z_1=z_0^*$$, $$z_2=1/z_0$$, $$z_3=1/z_0^*$$, $$z_4=3$$, $$z_5=1/z_4$$, $$z_6=-1$$. If we assume odd symmetry (type IV), the zeros are $$z_0=0.4e^{j\pi/3}$$, $$z_1=z_0^*$$, $$z_2=1/z_0$$, $$z_3=1/z_0^*$$, $$z_4=3$$, $$z_5=1/z_4$$, $$z_6=1$$. The corresponding impulse responses are $$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$ $$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$ If the problem description is correct, i.e., if $$h[n]$$ is zero for $$n<0$$ and $$n>7$$, and if we assume that $$h[7]\neq 0$$ (and why should we assume otherwise?), then the filter has $$8$$ taps and it is a $$7^{th}$$ order FIR filter with $$7$$ zeros. Since it is a linear phase filter and the number of taps is even, it must be either a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $$z=-1$$, in the second case it must have a zero at $$z=1$$. Refer to this answer for more details on the $$4$$ types of linear phase FIR filters. So if we assume even symmetry (type II), the zeros are $$z_0=0.4e^{j\pi/3}$$, $$z_1=z_0^*$$, $$z_2=1/z_0$$, $$z_3=1/z_0^*$$, $$z_4=3$$, $$z_5=1/z_4$$, $$z_6=-1$$. If we assume odd symmetry (type IV), the zeros are $$z_0=0.4e^{j\pi/3}$$, $$z_1=z_0^*$$, $$z_2=1/z_0$$, $$z_3=1/z_0^*$$, $$z_4=3$$, $$z_5=1/z_4$$, $$z_6=1$$. The corresponding impulse responses are $$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$ $$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$ EDIT: The answer above is correct (as far as I can tell) for the given problem description. Now that I've checked the original source (Schaum's Outlines of DSP), which includes the solution, I believe that there are the following possibilities: either there's a typo in the problem description and they actually meant that $$h[n]$$ is zero for $$n<0$$ and $$n\ge 7$$ (note the "greater or equal" sign). In this case we really have a $$6^{th}$$ order FIR filter and the given solution in Schaum's Outline is correct. The other option would be that they don't know what they're talking about (and I hope and believe that this is not the case). 3 edited body edited Jan 16 at 11:19 Matt L. 52.4k22 gold badges3939 silver badges9898 bronze badges If the problem description is correct, i.e., if $$h[n]$$ is zero for $$n<0$$ and $$n>7$$, and if we assume that $$h[7]\neq 0$$ (and why should we assume otherwise?), then the filter has $$8$$ taps and it is a $$7^{th}$$ order FIR filter with $$7$$ zeros. Since it is a linear phase filter and the number of taps is even, it must either be either a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $$z=-1$$, in the second case it must have a zero at $$z=1$$. Refer to this answer for more details on the $$4$$ types of linear phase FIR filters. So if we assume even symmetry (type II), the zeros are $$z_1=0.4e^{j\pi/3}$$$$z_0=0.4e^{j\pi/3}$$, $$z_1^*$$$$z_1=z_0^*$$, $$1/z_1$$$$z_2=1/z_0$$, $$1/z_1^*$$$$z_3=1/z_0^*$$, $$z_2=3$$$$z_4=3$$, $$1/z_2$$$$z_5=1/z_4$$, $$z_3=-1$$$$z_6=-1$$.  If we assume odd symmetry (type IV), the zeros are $$z_1=0.4e^{j\pi/3}$$$$z_0=0.4e^{j\pi/3}$$, $$z_1^*$$$$z_1=z_0^*$$, $$1/z_1$$$$z_2=1/z_0$$, $$1/z_1^*$$$$z_3=1/z_0^*$$, $$z_2=3$$$$z_4=3$$, $$1/z_2$$$$z_5=1/z_4$$, $$z_3=1$$$$z_6=1$$. The corresponding impulse responses are $$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$ $$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$ If the problem description is correct, i.e., if $$h[n]$$ is zero for $$n<0$$ and $$n>7$$, and if we assume that $$h[7]\neq 0$$ (and why should we assume otherwise?), then the filter has $$8$$ taps and it is a $$7^{th}$$ order FIR filter with $$7$$ zeros. Since it is a linear phase filter and the number of taps is even, it must either be a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $$z=-1$$, in the second case it must have a zero at $$z=1$$. So if we assume even symmetry (type II), the zeros are $$z_1=0.4e^{j\pi/3}$$, $$z_1^*$$, $$1/z_1$$, $$1/z_1^*$$, $$z_2=3$$, $$1/z_2$$, $$z_3=-1$$. If we assume odd symmetry (type IV), the zeros are $$z_1=0.4e^{j\pi/3}$$, $$z_1^*$$, $$1/z_1$$, $$1/z_1^*$$, $$z_2=3$$, $$1/z_2$$, $$z_3=1$$. The corresponding impulse responses are $$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$ $$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$ If the problem description is correct, i.e., if $$h[n]$$ is zero for $$n<0$$ and $$n>7$$, and if we assume that $$h[7]\neq 0$$ (and why should we assume otherwise?), then the filter has $$8$$ taps and it is a $$7^{th}$$ order FIR filter with $$7$$ zeros. Since it is a linear phase filter and the number of taps is even, it must be either a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $$z=-1$$, in the second case it must have a zero at $$z=1$$. Refer to this answer for more details on the $$4$$ types of linear phase FIR filters. So if we assume even symmetry (type II), the zeros are $$z_0=0.4e^{j\pi/3}$$, $$z_1=z_0^*$$, $$z_2=1/z_0$$, $$z_3=1/z_0^*$$, $$z_4=3$$, $$z_5=1/z_4$$, $$z_6=-1$$.  If we assume odd symmetry (type IV), the zeros are $$z_0=0.4e^{j\pi/3}$$, $$z_1=z_0^*$$, $$z_2=1/z_0$$, $$z_3=1/z_0^*$$, $$z_4=3$$, $$z_5=1/z_4$$, $$z_6=1$$. The corresponding impulse responses are $$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$ $$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$ 2 added 495 characters in body edited Jan 16 at 8:08 Matt L. 52.4k22 gold badges3939 silver badges9898 bronze badges If the problem description is correct, i.e., if $$h[n]$$ is non-zerozero for $$n=0,\ldots,7$$$$n<0$$ and $$n>7$$, and if we assume that $$h[7]\neq 0$$ (and why should we assume otherwise?), then itthe filter has $$8$$ taps and it is a $$7^{th}$$ order FIR filter with $$7$$ zeros. Since it is a linear phase filter and the number of taps is even, it must either be a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $$z=-1$$, in the second case it must have a zero at $$z=1$$. So if we assume even symmetry (type II), the zeros are $$z_1=0.4e^{j\pi/3}$$, $$z_1^*$$, $$1/z_1$$, $$1/z_1^*$$, $$z_2=3$$, $$1/z_2$$, $$z_3=-1$$. If we assume odd symmetry (type IV), the zeros are $$z_1=0.4e^{j\pi/3}$$, $$z_1^*$$, $$1/z_1$$, $$1/z_1^*$$, $$z_2=3$$, $$1/z_2$$, $$z_3=1$$. The corresponding impulse responses are $$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$ $$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$ If the problem description is correct, i.e., if $$h[n]$$ is non-zero for $$n=0,\ldots,7$$, then it has $$8$$ taps and it is a $$7^{th}$$ order FIR filter with $$7$$ zeros. Since the number of taps is even, it must either be a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $$z=-1$$, in the second case it must have a zero at $$z=1$$. If the problem description is correct, i.e., if $$h[n]$$ is zero for $$n<0$$ and $$n>7$$, and if we assume that $$h[7]\neq 0$$ (and why should we assume otherwise?), then the filter has $$8$$ taps and it is a $$7^{th}$$ order FIR filter with $$7$$ zeros. Since it is a linear phase filter and the number of taps is even, it must either be a type II filter (even number of taps, even symmetry), or a type IV filter (even number of taps, odd symmetry). In the first case it must have an additional zero at $$z=-1$$, in the second case it must have a zero at $$z=1$$. So if we assume even symmetry (type II), the zeros are $$z_1=0.4e^{j\pi/3}$$, $$z_1^*$$, $$1/z_1$$, $$1/z_1^*$$, $$z_2=3$$, $$1/z_2$$, $$z_3=-1$$. If we assume odd symmetry (type IV), the zeros are $$z_1=0.4e^{j\pi/3}$$, $$z_1^*$$, $$1/z_1$$, $$1/z_1^*$$, $$z_2=3$$, $$1/z_2$$, $$z_3=1$$. The corresponding impulse responses are $$h_{II}=[1, -5.23, 11.84, -12.42, -12.42, 11.84, -5.23, 1]$$ $$h_{IV}=[1, -7.23, 24.31, -48.57, 48.57, -24.31, 7.23, -1]$$ 1 answered Jan 16 at 8:03 Matt L. 52.4k22 gold badges3939 silver badges9898 bronze badges